Simple proof Euler–Mascheroni $gamma$ constant












11














I'm searching for a really simple and beautiful proof that the sequence $(u_n)_{n in mathbb{N}} = sumnolimits_{k=1}^n frac{1}{k} - log(n)$ converges.

At first I want to know if my answer is OK.



My try:

$limlimits_{ntoinfty} sumlimits_{k=1}^n frac{1}{k} - log (n) = limlimits_{ntoinfty} sumlimits_{k=1}^n frac{1}{k} + sumlimits_{k=1}^{n-1} log(k)-log(k+1) = limlimits_{ntoinfty} frac{1}{n} + sumlimits_{k=1}^{n-1} log(frac{k}{k+1})+frac{1}{k}$



$ = sumlimits_{k=1}^{infty} frac{1}{k}-log(frac{k+1}{k})$
Now we prove that the last sum converges by the comparison test:

$frac{1}{k}-log(frac{k+1}{k}) < frac{1}{k^2} Leftrightarrow k<k^2log(frac{k+1}{k})+1$

which surely holds for $kgeqslant 1$





As $ sumlimits_{k=1}^{infty} frac{1}{k^2}$ converges $ Rightarrow sumlimits_{k=1}^{infty} frac{1}{k}-log(frac{k+1}{k})$ converges and we name this limit $gamma$

q.e.d










share|cite|improve this question




















  • 1




    Hint: look at the function $y=1/x$ compared to "step functions" above and below that curve, and look at what happens when you integrate.
    – Old John
    Jan 6 '14 at 23:59






  • 1




    math.stackexchange.com/questions/306371/…
    – Will Jagy
    Jan 7 '14 at 0:01






  • 1




    the diagrams are not very good, but you can get the idea here.
    – Old John
    Jan 7 '14 at 0:05






  • 1




    @OldJohn, I see, actually a link to a Macalester College discussion of the harmonic series. Natural mistake.
    – Will Jagy
    Jan 7 '14 at 0:09






  • 2




    @OldJohn, true, never met her. There is a running joke on a TV series called NCIS, in which FBI agent Fornell married a woman whom NCIS agent Gibbs had divorced. At one point Fornell says "In my defense, I didn't believe him."
    – Will Jagy
    Jan 7 '14 at 0:28


















11














I'm searching for a really simple and beautiful proof that the sequence $(u_n)_{n in mathbb{N}} = sumnolimits_{k=1}^n frac{1}{k} - log(n)$ converges.

At first I want to know if my answer is OK.



My try:

$limlimits_{ntoinfty} sumlimits_{k=1}^n frac{1}{k} - log (n) = limlimits_{ntoinfty} sumlimits_{k=1}^n frac{1}{k} + sumlimits_{k=1}^{n-1} log(k)-log(k+1) = limlimits_{ntoinfty} frac{1}{n} + sumlimits_{k=1}^{n-1} log(frac{k}{k+1})+frac{1}{k}$



$ = sumlimits_{k=1}^{infty} frac{1}{k}-log(frac{k+1}{k})$
Now we prove that the last sum converges by the comparison test:

$frac{1}{k}-log(frac{k+1}{k}) < frac{1}{k^2} Leftrightarrow k<k^2log(frac{k+1}{k})+1$

which surely holds for $kgeqslant 1$





As $ sumlimits_{k=1}^{infty} frac{1}{k^2}$ converges $ Rightarrow sumlimits_{k=1}^{infty} frac{1}{k}-log(frac{k+1}{k})$ converges and we name this limit $gamma$

q.e.d










share|cite|improve this question




















  • 1




    Hint: look at the function $y=1/x$ compared to "step functions" above and below that curve, and look at what happens when you integrate.
    – Old John
    Jan 6 '14 at 23:59






  • 1




    math.stackexchange.com/questions/306371/…
    – Will Jagy
    Jan 7 '14 at 0:01






  • 1




    the diagrams are not very good, but you can get the idea here.
    – Old John
    Jan 7 '14 at 0:05






  • 1




    @OldJohn, I see, actually a link to a Macalester College discussion of the harmonic series. Natural mistake.
    – Will Jagy
    Jan 7 '14 at 0:09






  • 2




    @OldJohn, true, never met her. There is a running joke on a TV series called NCIS, in which FBI agent Fornell married a woman whom NCIS agent Gibbs had divorced. At one point Fornell says "In my defense, I didn't believe him."
    – Will Jagy
    Jan 7 '14 at 0:28
















11












11








11


5





I'm searching for a really simple and beautiful proof that the sequence $(u_n)_{n in mathbb{N}} = sumnolimits_{k=1}^n frac{1}{k} - log(n)$ converges.

At first I want to know if my answer is OK.



My try:

$limlimits_{ntoinfty} sumlimits_{k=1}^n frac{1}{k} - log (n) = limlimits_{ntoinfty} sumlimits_{k=1}^n frac{1}{k} + sumlimits_{k=1}^{n-1} log(k)-log(k+1) = limlimits_{ntoinfty} frac{1}{n} + sumlimits_{k=1}^{n-1} log(frac{k}{k+1})+frac{1}{k}$



$ = sumlimits_{k=1}^{infty} frac{1}{k}-log(frac{k+1}{k})$
Now we prove that the last sum converges by the comparison test:

$frac{1}{k}-log(frac{k+1}{k}) < frac{1}{k^2} Leftrightarrow k<k^2log(frac{k+1}{k})+1$

which surely holds for $kgeqslant 1$





As $ sumlimits_{k=1}^{infty} frac{1}{k^2}$ converges $ Rightarrow sumlimits_{k=1}^{infty} frac{1}{k}-log(frac{k+1}{k})$ converges and we name this limit $gamma$

q.e.d










share|cite|improve this question















I'm searching for a really simple and beautiful proof that the sequence $(u_n)_{n in mathbb{N}} = sumnolimits_{k=1}^n frac{1}{k} - log(n)$ converges.

At first I want to know if my answer is OK.



My try:

$limlimits_{ntoinfty} sumlimits_{k=1}^n frac{1}{k} - log (n) = limlimits_{ntoinfty} sumlimits_{k=1}^n frac{1}{k} + sumlimits_{k=1}^{n-1} log(k)-log(k+1) = limlimits_{ntoinfty} frac{1}{n} + sumlimits_{k=1}^{n-1} log(frac{k}{k+1})+frac{1}{k}$



$ = sumlimits_{k=1}^{infty} frac{1}{k}-log(frac{k+1}{k})$
Now we prove that the last sum converges by the comparison test:

$frac{1}{k}-log(frac{k+1}{k}) < frac{1}{k^2} Leftrightarrow k<k^2log(frac{k+1}{k})+1$

which surely holds for $kgeqslant 1$





As $ sumlimits_{k=1}^{infty} frac{1}{k^2}$ converges $ Rightarrow sumlimits_{k=1}^{infty} frac{1}{k}-log(frac{k+1}{k})$ converges and we name this limit $gamma$

q.e.d







limits logarithms eulers-constant harmonic-numbers






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share|cite|improve this question













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share|cite|improve this question








edited Apr 27 '17 at 6:53









k.Vijay

1,927317




1,927317










asked Jan 6 '14 at 23:55









Thomas Produit

6351721




6351721








  • 1




    Hint: look at the function $y=1/x$ compared to "step functions" above and below that curve, and look at what happens when you integrate.
    – Old John
    Jan 6 '14 at 23:59






  • 1




    math.stackexchange.com/questions/306371/…
    – Will Jagy
    Jan 7 '14 at 0:01






  • 1




    the diagrams are not very good, but you can get the idea here.
    – Old John
    Jan 7 '14 at 0:05






  • 1




    @OldJohn, I see, actually a link to a Macalester College discussion of the harmonic series. Natural mistake.
    – Will Jagy
    Jan 7 '14 at 0:09






  • 2




    @OldJohn, true, never met her. There is a running joke on a TV series called NCIS, in which FBI agent Fornell married a woman whom NCIS agent Gibbs had divorced. At one point Fornell says "In my defense, I didn't believe him."
    – Will Jagy
    Jan 7 '14 at 0:28
















  • 1




    Hint: look at the function $y=1/x$ compared to "step functions" above and below that curve, and look at what happens when you integrate.
    – Old John
    Jan 6 '14 at 23:59






  • 1




    math.stackexchange.com/questions/306371/…
    – Will Jagy
    Jan 7 '14 at 0:01






  • 1




    the diagrams are not very good, but you can get the idea here.
    – Old John
    Jan 7 '14 at 0:05






  • 1




    @OldJohn, I see, actually a link to a Macalester College discussion of the harmonic series. Natural mistake.
    – Will Jagy
    Jan 7 '14 at 0:09






  • 2




    @OldJohn, true, never met her. There is a running joke on a TV series called NCIS, in which FBI agent Fornell married a woman whom NCIS agent Gibbs had divorced. At one point Fornell says "In my defense, I didn't believe him."
    – Will Jagy
    Jan 7 '14 at 0:28










1




1




Hint: look at the function $y=1/x$ compared to "step functions" above and below that curve, and look at what happens when you integrate.
– Old John
Jan 6 '14 at 23:59




Hint: look at the function $y=1/x$ compared to "step functions" above and below that curve, and look at what happens when you integrate.
– Old John
Jan 6 '14 at 23:59




1




1




math.stackexchange.com/questions/306371/…
– Will Jagy
Jan 7 '14 at 0:01




math.stackexchange.com/questions/306371/…
– Will Jagy
Jan 7 '14 at 0:01




1




1




the diagrams are not very good, but you can get the idea here.
– Old John
Jan 7 '14 at 0:05




the diagrams are not very good, but you can get the idea here.
– Old John
Jan 7 '14 at 0:05




1




1




@OldJohn, I see, actually a link to a Macalester College discussion of the harmonic series. Natural mistake.
– Will Jagy
Jan 7 '14 at 0:09




@OldJohn, I see, actually a link to a Macalester College discussion of the harmonic series. Natural mistake.
– Will Jagy
Jan 7 '14 at 0:09




2




2




@OldJohn, true, never met her. There is a running joke on a TV series called NCIS, in which FBI agent Fornell married a woman whom NCIS agent Gibbs had divorced. At one point Fornell says "In my defense, I didn't believe him."
– Will Jagy
Jan 7 '14 at 0:28






@OldJohn, true, never met her. There is a running joke on a TV series called NCIS, in which FBI agent Fornell married a woman whom NCIS agent Gibbs had divorced. At one point Fornell says "In my defense, I didn't believe him."
– Will Jagy
Jan 7 '14 at 0:28












2 Answers
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oldest

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13














One elegant way to show that the sequence converges is to show that it's both decreasing and bounded below.



It's decreasing because $u_n-u_{n-1} = frac1n - log n + log(n-1) = frac1n + log(1-frac1n) < 0$ for all $n$. (The inequality is valid because $log(1-x)$ is a concave function, hence lies beneath the line $-x$ that is tangent to its graph at $0$; plugging in $x=frac1n$ yields $log(1-frac1n) le -frac1n$.)



It's bounded below because
$$
sum_{j=1}^n frac1j > int_1^{n+1} frac{dt}t = log (n+1) > log n,
$$
and so $u_n>0$ for all $n$. (The inequality is valid because the sum is a left-hand endpoint Riemann sum for the integral, and the function $frac1t$ is decreasing.)






share|cite|improve this answer































    3














    Upper Bound



    Note that
    $$
    begin{align}
    frac1n-logleft(frac{n+1}nright)
    &=int_0^{1/n}frac{t,mathrm{d}t}{1+t}\
    &leint_0^{1/n}t,mathrm{d}t\[3pt]
    &=frac1{2n^2}
    end{align}
    $$

    Therefore,
    $$
    begin{align}
    gamma
    &=sum_{n=1}^inftyleft(frac1n-logleft(frac{n+1}nright)right)\
    &lesum_{n=1}^inftyfrac1{2n^2}\
    &lesum_{n=1}^inftyfrac1{2n^2-frac12}\
    &=sum_{n=1}^inftyfrac12left(frac1{n-frac12}-frac1{n+frac12}right)\[9pt]
    &=1
    end{align}
    $$





    Lower Bound



    Note that
    $$
    begin{align}
    frac1n-logleft(frac{n+1}nright)
    &=int_0^{1/n}frac{t,mathrm{d}t}{1+t}\
    &geint_0^{1/n}frac{t}{1+frac1n},mathrm{d}t\[3pt]
    &=frac1{2n(n+1)}
    end{align}
    $$

    Therefore,
    $$
    begin{align}
    gamma
    &=sum_{n=1}^inftyleft(frac1n-logleft(frac{n+1}nright)right)\
    &gesum_{n=1}^inftyfrac1{2n(n+1)}\[3pt]
    &=sum_{n=1}^inftyfrac12left(frac1n-frac1{n+1}right)\[6pt]
    &=frac12
    end{align}
    $$





    A Better Upper Bound



    Using Jensen's Inequality on the concave $frac{t}{1+t}$, we get
    $$
    begin{align}
    frac1n-logleft(frac{n+1}nright)
    &=frac1nleft(nint_0^{1/n}frac{t,mathrm{d}t}{1+t}right)\
    &lefrac1nfrac{nint_0^{1/n}t,mathrm{d}t}{1+nint_0^{1/n}t,mathrm{d}t}\
    &=frac1{n(2n+1)}
    end{align}
    $$

    Therefore, since the sum of the Alternating Harmonic Series is $log(2)$,
    $$
    begin{align}
    gamma
    &=sum_{n=1}^inftyleft(frac1n-logleft(frac{n+1}nright)right)\
    &lesum_{n=1}^inftyfrac1{n(2n+1)}\
    &=sum_{n=1}^infty2left(frac1{2n}-frac1{2n+1}right)\[6pt]
    &=2(1-log(2))
    end{align}
    $$






    share|cite|improve this answer























    • can you copy the last part of your answer here ?
      – Gabriel Romon
      Nov 10 '17 at 9:52













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

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    active

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    active

    oldest

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    13














    One elegant way to show that the sequence converges is to show that it's both decreasing and bounded below.



    It's decreasing because $u_n-u_{n-1} = frac1n - log n + log(n-1) = frac1n + log(1-frac1n) < 0$ for all $n$. (The inequality is valid because $log(1-x)$ is a concave function, hence lies beneath the line $-x$ that is tangent to its graph at $0$; plugging in $x=frac1n$ yields $log(1-frac1n) le -frac1n$.)



    It's bounded below because
    $$
    sum_{j=1}^n frac1j > int_1^{n+1} frac{dt}t = log (n+1) > log n,
    $$
    and so $u_n>0$ for all $n$. (The inequality is valid because the sum is a left-hand endpoint Riemann sum for the integral, and the function $frac1t$ is decreasing.)






    share|cite|improve this answer




























      13














      One elegant way to show that the sequence converges is to show that it's both decreasing and bounded below.



      It's decreasing because $u_n-u_{n-1} = frac1n - log n + log(n-1) = frac1n + log(1-frac1n) < 0$ for all $n$. (The inequality is valid because $log(1-x)$ is a concave function, hence lies beneath the line $-x$ that is tangent to its graph at $0$; plugging in $x=frac1n$ yields $log(1-frac1n) le -frac1n$.)



      It's bounded below because
      $$
      sum_{j=1}^n frac1j > int_1^{n+1} frac{dt}t = log (n+1) > log n,
      $$
      and so $u_n>0$ for all $n$. (The inequality is valid because the sum is a left-hand endpoint Riemann sum for the integral, and the function $frac1t$ is decreasing.)






      share|cite|improve this answer


























        13












        13








        13






        One elegant way to show that the sequence converges is to show that it's both decreasing and bounded below.



        It's decreasing because $u_n-u_{n-1} = frac1n - log n + log(n-1) = frac1n + log(1-frac1n) < 0$ for all $n$. (The inequality is valid because $log(1-x)$ is a concave function, hence lies beneath the line $-x$ that is tangent to its graph at $0$; plugging in $x=frac1n$ yields $log(1-frac1n) le -frac1n$.)



        It's bounded below because
        $$
        sum_{j=1}^n frac1j > int_1^{n+1} frac{dt}t = log (n+1) > log n,
        $$
        and so $u_n>0$ for all $n$. (The inequality is valid because the sum is a left-hand endpoint Riemann sum for the integral, and the function $frac1t$ is decreasing.)






        share|cite|improve this answer














        One elegant way to show that the sequence converges is to show that it's both decreasing and bounded below.



        It's decreasing because $u_n-u_{n-1} = frac1n - log n + log(n-1) = frac1n + log(1-frac1n) < 0$ for all $n$. (The inequality is valid because $log(1-x)$ is a concave function, hence lies beneath the line $-x$ that is tangent to its graph at $0$; plugging in $x=frac1n$ yields $log(1-frac1n) le -frac1n$.)



        It's bounded below because
        $$
        sum_{j=1}^n frac1j > int_1^{n+1} frac{dt}t = log (n+1) > log n,
        $$
        and so $u_n>0$ for all $n$. (The inequality is valid because the sum is a left-hand endpoint Riemann sum for the integral, and the function $frac1t$ is decreasing.)







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jul 5 '14 at 22:57









        hardmath

        28.7k94994




        28.7k94994










        answered Jan 7 '14 at 1:07









        Greg Martin

        34.7k23161




        34.7k23161























            3














            Upper Bound



            Note that
            $$
            begin{align}
            frac1n-logleft(frac{n+1}nright)
            &=int_0^{1/n}frac{t,mathrm{d}t}{1+t}\
            &leint_0^{1/n}t,mathrm{d}t\[3pt]
            &=frac1{2n^2}
            end{align}
            $$

            Therefore,
            $$
            begin{align}
            gamma
            &=sum_{n=1}^inftyleft(frac1n-logleft(frac{n+1}nright)right)\
            &lesum_{n=1}^inftyfrac1{2n^2}\
            &lesum_{n=1}^inftyfrac1{2n^2-frac12}\
            &=sum_{n=1}^inftyfrac12left(frac1{n-frac12}-frac1{n+frac12}right)\[9pt]
            &=1
            end{align}
            $$





            Lower Bound



            Note that
            $$
            begin{align}
            frac1n-logleft(frac{n+1}nright)
            &=int_0^{1/n}frac{t,mathrm{d}t}{1+t}\
            &geint_0^{1/n}frac{t}{1+frac1n},mathrm{d}t\[3pt]
            &=frac1{2n(n+1)}
            end{align}
            $$

            Therefore,
            $$
            begin{align}
            gamma
            &=sum_{n=1}^inftyleft(frac1n-logleft(frac{n+1}nright)right)\
            &gesum_{n=1}^inftyfrac1{2n(n+1)}\[3pt]
            &=sum_{n=1}^inftyfrac12left(frac1n-frac1{n+1}right)\[6pt]
            &=frac12
            end{align}
            $$





            A Better Upper Bound



            Using Jensen's Inequality on the concave $frac{t}{1+t}$, we get
            $$
            begin{align}
            frac1n-logleft(frac{n+1}nright)
            &=frac1nleft(nint_0^{1/n}frac{t,mathrm{d}t}{1+t}right)\
            &lefrac1nfrac{nint_0^{1/n}t,mathrm{d}t}{1+nint_0^{1/n}t,mathrm{d}t}\
            &=frac1{n(2n+1)}
            end{align}
            $$

            Therefore, since the sum of the Alternating Harmonic Series is $log(2)$,
            $$
            begin{align}
            gamma
            &=sum_{n=1}^inftyleft(frac1n-logleft(frac{n+1}nright)right)\
            &lesum_{n=1}^inftyfrac1{n(2n+1)}\
            &=sum_{n=1}^infty2left(frac1{2n}-frac1{2n+1}right)\[6pt]
            &=2(1-log(2))
            end{align}
            $$






            share|cite|improve this answer























            • can you copy the last part of your answer here ?
              – Gabriel Romon
              Nov 10 '17 at 9:52


















            3














            Upper Bound



            Note that
            $$
            begin{align}
            frac1n-logleft(frac{n+1}nright)
            &=int_0^{1/n}frac{t,mathrm{d}t}{1+t}\
            &leint_0^{1/n}t,mathrm{d}t\[3pt]
            &=frac1{2n^2}
            end{align}
            $$

            Therefore,
            $$
            begin{align}
            gamma
            &=sum_{n=1}^inftyleft(frac1n-logleft(frac{n+1}nright)right)\
            &lesum_{n=1}^inftyfrac1{2n^2}\
            &lesum_{n=1}^inftyfrac1{2n^2-frac12}\
            &=sum_{n=1}^inftyfrac12left(frac1{n-frac12}-frac1{n+frac12}right)\[9pt]
            &=1
            end{align}
            $$





            Lower Bound



            Note that
            $$
            begin{align}
            frac1n-logleft(frac{n+1}nright)
            &=int_0^{1/n}frac{t,mathrm{d}t}{1+t}\
            &geint_0^{1/n}frac{t}{1+frac1n},mathrm{d}t\[3pt]
            &=frac1{2n(n+1)}
            end{align}
            $$

            Therefore,
            $$
            begin{align}
            gamma
            &=sum_{n=1}^inftyleft(frac1n-logleft(frac{n+1}nright)right)\
            &gesum_{n=1}^inftyfrac1{2n(n+1)}\[3pt]
            &=sum_{n=1}^inftyfrac12left(frac1n-frac1{n+1}right)\[6pt]
            &=frac12
            end{align}
            $$





            A Better Upper Bound



            Using Jensen's Inequality on the concave $frac{t}{1+t}$, we get
            $$
            begin{align}
            frac1n-logleft(frac{n+1}nright)
            &=frac1nleft(nint_0^{1/n}frac{t,mathrm{d}t}{1+t}right)\
            &lefrac1nfrac{nint_0^{1/n}t,mathrm{d}t}{1+nint_0^{1/n}t,mathrm{d}t}\
            &=frac1{n(2n+1)}
            end{align}
            $$

            Therefore, since the sum of the Alternating Harmonic Series is $log(2)$,
            $$
            begin{align}
            gamma
            &=sum_{n=1}^inftyleft(frac1n-logleft(frac{n+1}nright)right)\
            &lesum_{n=1}^inftyfrac1{n(2n+1)}\
            &=sum_{n=1}^infty2left(frac1{2n}-frac1{2n+1}right)\[6pt]
            &=2(1-log(2))
            end{align}
            $$






            share|cite|improve this answer























            • can you copy the last part of your answer here ?
              – Gabriel Romon
              Nov 10 '17 at 9:52
















            3












            3








            3






            Upper Bound



            Note that
            $$
            begin{align}
            frac1n-logleft(frac{n+1}nright)
            &=int_0^{1/n}frac{t,mathrm{d}t}{1+t}\
            &leint_0^{1/n}t,mathrm{d}t\[3pt]
            &=frac1{2n^2}
            end{align}
            $$

            Therefore,
            $$
            begin{align}
            gamma
            &=sum_{n=1}^inftyleft(frac1n-logleft(frac{n+1}nright)right)\
            &lesum_{n=1}^inftyfrac1{2n^2}\
            &lesum_{n=1}^inftyfrac1{2n^2-frac12}\
            &=sum_{n=1}^inftyfrac12left(frac1{n-frac12}-frac1{n+frac12}right)\[9pt]
            &=1
            end{align}
            $$





            Lower Bound



            Note that
            $$
            begin{align}
            frac1n-logleft(frac{n+1}nright)
            &=int_0^{1/n}frac{t,mathrm{d}t}{1+t}\
            &geint_0^{1/n}frac{t}{1+frac1n},mathrm{d}t\[3pt]
            &=frac1{2n(n+1)}
            end{align}
            $$

            Therefore,
            $$
            begin{align}
            gamma
            &=sum_{n=1}^inftyleft(frac1n-logleft(frac{n+1}nright)right)\
            &gesum_{n=1}^inftyfrac1{2n(n+1)}\[3pt]
            &=sum_{n=1}^inftyfrac12left(frac1n-frac1{n+1}right)\[6pt]
            &=frac12
            end{align}
            $$





            A Better Upper Bound



            Using Jensen's Inequality on the concave $frac{t}{1+t}$, we get
            $$
            begin{align}
            frac1n-logleft(frac{n+1}nright)
            &=frac1nleft(nint_0^{1/n}frac{t,mathrm{d}t}{1+t}right)\
            &lefrac1nfrac{nint_0^{1/n}t,mathrm{d}t}{1+nint_0^{1/n}t,mathrm{d}t}\
            &=frac1{n(2n+1)}
            end{align}
            $$

            Therefore, since the sum of the Alternating Harmonic Series is $log(2)$,
            $$
            begin{align}
            gamma
            &=sum_{n=1}^inftyleft(frac1n-logleft(frac{n+1}nright)right)\
            &lesum_{n=1}^inftyfrac1{n(2n+1)}\
            &=sum_{n=1}^infty2left(frac1{2n}-frac1{2n+1}right)\[6pt]
            &=2(1-log(2))
            end{align}
            $$






            share|cite|improve this answer














            Upper Bound



            Note that
            $$
            begin{align}
            frac1n-logleft(frac{n+1}nright)
            &=int_0^{1/n}frac{t,mathrm{d}t}{1+t}\
            &leint_0^{1/n}t,mathrm{d}t\[3pt]
            &=frac1{2n^2}
            end{align}
            $$

            Therefore,
            $$
            begin{align}
            gamma
            &=sum_{n=1}^inftyleft(frac1n-logleft(frac{n+1}nright)right)\
            &lesum_{n=1}^inftyfrac1{2n^2}\
            &lesum_{n=1}^inftyfrac1{2n^2-frac12}\
            &=sum_{n=1}^inftyfrac12left(frac1{n-frac12}-frac1{n+frac12}right)\[9pt]
            &=1
            end{align}
            $$





            Lower Bound



            Note that
            $$
            begin{align}
            frac1n-logleft(frac{n+1}nright)
            &=int_0^{1/n}frac{t,mathrm{d}t}{1+t}\
            &geint_0^{1/n}frac{t}{1+frac1n},mathrm{d}t\[3pt]
            &=frac1{2n(n+1)}
            end{align}
            $$

            Therefore,
            $$
            begin{align}
            gamma
            &=sum_{n=1}^inftyleft(frac1n-logleft(frac{n+1}nright)right)\
            &gesum_{n=1}^inftyfrac1{2n(n+1)}\[3pt]
            &=sum_{n=1}^inftyfrac12left(frac1n-frac1{n+1}right)\[6pt]
            &=frac12
            end{align}
            $$





            A Better Upper Bound



            Using Jensen's Inequality on the concave $frac{t}{1+t}$, we get
            $$
            begin{align}
            frac1n-logleft(frac{n+1}nright)
            &=frac1nleft(nint_0^{1/n}frac{t,mathrm{d}t}{1+t}right)\
            &lefrac1nfrac{nint_0^{1/n}t,mathrm{d}t}{1+nint_0^{1/n}t,mathrm{d}t}\
            &=frac1{n(2n+1)}
            end{align}
            $$

            Therefore, since the sum of the Alternating Harmonic Series is $log(2)$,
            $$
            begin{align}
            gamma
            &=sum_{n=1}^inftyleft(frac1n-logleft(frac{n+1}nright)right)\
            &lesum_{n=1}^inftyfrac1{n(2n+1)}\
            &=sum_{n=1}^infty2left(frac1{2n}-frac1{2n+1}right)\[6pt]
            &=2(1-log(2))
            end{align}
            $$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 1 hour ago

























            answered Sep 24 '17 at 12:52









            robjohn

            264k27302623




            264k27302623












            • can you copy the last part of your answer here ?
              – Gabriel Romon
              Nov 10 '17 at 9:52




















            • can you copy the last part of your answer here ?
              – Gabriel Romon
              Nov 10 '17 at 9:52


















            can you copy the last part of your answer here ?
            – Gabriel Romon
            Nov 10 '17 at 9:52






            can you copy the last part of your answer here ?
            – Gabriel Romon
            Nov 10 '17 at 9:52




















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