Dtyjlui

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

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.)

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}
$$