Does $2sumfrac{1}{(2k+1)3^{2k+1}}$ converge to $ln 2$ linearly?












0












$begingroup$


Based on the fact that
$$frac{1}{2}lnfrac{1+x}{1-x}=sum_{k=0}^inftyfrac{x^{2k+1}}{2k+1}$$
and evaluating at $x=-1/3$, we can conclude
$$ln 2=2sum_{k=0}^infty frac{1}{(2k+1)3^{2k+1}},.$$




However, I'm interested in the speed of this convergence. Is it true that this series converges to the limit linearly in the sense that
$$limsup_{nrightarrowinfty}frac{|ln 2-S_{n+1}|}{|ln 2-S_{n}|}<1$$
where ${S_n}$ are the partial sums?




It's pretty straight-forward to calculate
$$|ln 2-S_n|=sum_{k=n+1}^inftyfrac{1}{(2k+1)3^{2k+1}}<frac{1}{2n+3}sum_{k=n+1}^inftyfrac{1}{3^{2k+1}}=frac{1}{8(6n+9)9^n},.$$
However, an effective lower bound on this tail is a little harder to find that'll make the ratio work out.






real-analysis sequences-and-series convergence rate-of-convergence







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$endgroup$












  • $begingroup$
    Why was this down-voted?...
    $endgroup$
    – Robert Wolfe
    Jan 21 at 23:02
















0












$begingroup$


Based on the fact that
$$frac{1}{2}lnfrac{1+x}{1-x}=sum_{k=0}^inftyfrac{x^{2k+1}}{2k+1}$$
and evaluating at $x=-1/3$, we can conclude
$$ln 2=2sum_{k=0}^infty frac{1}{(2k+1)3^{2k+1}},.$$




However, I'm interested in the speed of this convergence. Is it true that this series converges to the limit linearly in the sense that
$$limsup_{nrightarrowinfty}frac{|ln 2-S_{n+1}|}{|ln 2-S_{n}|}<1$$
where ${S_n}$ are the partial sums?




It's pretty straight-forward to calculate
$$|ln 2-S_n|=sum_{k=n+1}^inftyfrac{1}{(2k+1)3^{2k+1}}<frac{1}{2n+3}sum_{k=n+1}^inftyfrac{1}{3^{2k+1}}=frac{1}{8(6n+9)9^n},.$$
However, an effective lower bound on this tail is a little harder to find that'll make the ratio work out.






real-analysis sequences-and-series convergence rate-of-convergence







share|cite|improve this question









$endgroup$












  • $begingroup$
    Why was this down-voted?...
    $endgroup$
    – Robert Wolfe
    Jan 21 at 23:02














0












0








0





$begingroup$


Based on the fact that
$$frac{1}{2}lnfrac{1+x}{1-x}=sum_{k=0}^inftyfrac{x^{2k+1}}{2k+1}$$
and evaluating at $x=-1/3$, we can conclude
$$ln 2=2sum_{k=0}^infty frac{1}{(2k+1)3^{2k+1}},.$$




However, I'm interested in the speed of this convergence. Is it true that this series converges to the limit linearly in the sense that
$$limsup_{nrightarrowinfty}frac{|ln 2-S_{n+1}|}{|ln 2-S_{n}|}<1$$
where ${S_n}$ are the partial sums?




It's pretty straight-forward to calculate
$$|ln 2-S_n|=sum_{k=n+1}^inftyfrac{1}{(2k+1)3^{2k+1}}<frac{1}{2n+3}sum_{k=n+1}^inftyfrac{1}{3^{2k+1}}=frac{1}{8(6n+9)9^n},.$$
However, an effective lower bound on this tail is a little harder to find that'll make the ratio work out.






real-analysis sequences-and-series convergence rate-of-convergence







share|cite|improve this question









$endgroup$




Based on the fact that
$$frac{1}{2}lnfrac{1+x}{1-x}=sum_{k=0}^inftyfrac{x^{2k+1}}{2k+1}$$
and evaluating at $x=-1/3$, we can conclude
$$ln 2=2sum_{k=0}^infty frac{1}{(2k+1)3^{2k+1}},.$$




However, I'm interested in the speed of this convergence. Is it true that this series converges to the limit linearly in the sense that
$$limsup_{nrightarrowinfty}frac{|ln 2-S_{n+1}|}{|ln 2-S_{n}|}<1$$
where ${S_n}$ are the partial sums?




It's pretty straight-forward to calculate
$$|ln 2-S_n|=sum_{k=n+1}^inftyfrac{1}{(2k+1)3^{2k+1}}<frac{1}{2n+3}sum_{k=n+1}^inftyfrac{1}{3^{2k+1}}=frac{1}{8(6n+9)9^n},.$$
However, an effective lower bound on this tail is a little harder to find that'll make the ratio work out.






real-analysis sequences-and-series convergence rate-of-convergence




real-analysis sequences-and-series convergence rate-of-convergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 11 at 22:56









Robert WolfeRobert Wolfe

5,96922763




5,96922763












  • $begingroup$
    Why was this down-voted?...
    $endgroup$
    – Robert Wolfe
    Jan 21 at 23:02


















  • $begingroup$
    Why was this down-voted?...
    $endgroup$
    – Robert Wolfe
    Jan 21 at 23:02
















$begingroup$
Why was this down-voted?...
$endgroup$
– Robert Wolfe
Jan 21 at 23:02




$begingroup$
Why was this down-voted?...
$endgroup$
– Robert Wolfe
Jan 21 at 23:02










1 Answer
1






active

oldest

votes


















1












$begingroup$

We can use integral remainders:



begin{align}sum_{k=n}^inftyfrac1{(2k+1)3^{2k+1}}&=sum_{k=n}^inftyint_0^{1/3}x^{2k}~mathrm dx\&stackrel?=int_0^{1/3}sum_{k=n}^infty x^{2k}~mathrm dx\&=int_0^{1/3}frac{x^{2n+1}}{1-x^2}~mathrm dx\&>int_0^{1/3}frac{x^{2n+1}}{1-(1/3)^2}~mathrm dx\&=frac1{16(n+1)9^n}end{align}



which should be sufficient. Note this also proves the slightly tighter upper bound:



$$sum_{k=n}^inftyfrac1{(2k+1)3^{2k+1}}<int_0^{1/3}frac{x^{2n+1}}{1-0^2}~mathrm dx=frac1{18(n+1)9^n}$$



and hence you can see that



$$limsup_{ntoinfty}frac{|ln2-S_{n+1}|}{|ln2-S_n|}lefrac8{81}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That makes me wonder for what $alpha$ is the tail sequence equivalent to $alpha/(n9^n)$. Different question though, and something else for me try. Thanks!
    $endgroup$
    – Robert Wolfe
    Jan 12 at 1:19






  • 1




    $begingroup$
    @RobertWolfe Hint: integrate my last non-bound integral, by parts.
    $endgroup$
    – Simply Beautiful Art
    Jan 12 at 1:20













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

We can use integral remainders:



begin{align}sum_{k=n}^inftyfrac1{(2k+1)3^{2k+1}}&=sum_{k=n}^inftyint_0^{1/3}x^{2k}~mathrm dx\&stackrel?=int_0^{1/3}sum_{k=n}^infty x^{2k}~mathrm dx\&=int_0^{1/3}frac{x^{2n+1}}{1-x^2}~mathrm dx\&>int_0^{1/3}frac{x^{2n+1}}{1-(1/3)^2}~mathrm dx\&=frac1{16(n+1)9^n}end{align}



which should be sufficient. Note this also proves the slightly tighter upper bound:



$$sum_{k=n}^inftyfrac1{(2k+1)3^{2k+1}}<int_0^{1/3}frac{x^{2n+1}}{1-0^2}~mathrm dx=frac1{18(n+1)9^n}$$



and hence you can see that



$$limsup_{ntoinfty}frac{|ln2-S_{n+1}|}{|ln2-S_n|}lefrac8{81}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That makes me wonder for what $alpha$ is the tail sequence equivalent to $alpha/(n9^n)$. Different question though, and something else for me try. Thanks!
    $endgroup$
    – Robert Wolfe
    Jan 12 at 1:19






  • 1




    $begingroup$
    @RobertWolfe Hint: integrate my last non-bound integral, by parts.
    $endgroup$
    – Simply Beautiful Art
    Jan 12 at 1:20


















1












$begingroup$

We can use integral remainders:



begin{align}sum_{k=n}^inftyfrac1{(2k+1)3^{2k+1}}&=sum_{k=n}^inftyint_0^{1/3}x^{2k}~mathrm dx\&stackrel?=int_0^{1/3}sum_{k=n}^infty x^{2k}~mathrm dx\&=int_0^{1/3}frac{x^{2n+1}}{1-x^2}~mathrm dx\&>int_0^{1/3}frac{x^{2n+1}}{1-(1/3)^2}~mathrm dx\&=frac1{16(n+1)9^n}end{align}



which should be sufficient. Note this also proves the slightly tighter upper bound:



$$sum_{k=n}^inftyfrac1{(2k+1)3^{2k+1}}<int_0^{1/3}frac{x^{2n+1}}{1-0^2}~mathrm dx=frac1{18(n+1)9^n}$$



and hence you can see that



$$limsup_{ntoinfty}frac{|ln2-S_{n+1}|}{|ln2-S_n|}lefrac8{81}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That makes me wonder for what $alpha$ is the tail sequence equivalent to $alpha/(n9^n)$. Different question though, and something else for me try. Thanks!
    $endgroup$
    – Robert Wolfe
    Jan 12 at 1:19






  • 1




    $begingroup$
    @RobertWolfe Hint: integrate my last non-bound integral, by parts.
    $endgroup$
    – Simply Beautiful Art
    Jan 12 at 1:20
















1












1








1





$begingroup$

We can use integral remainders:



begin{align}sum_{k=n}^inftyfrac1{(2k+1)3^{2k+1}}&=sum_{k=n}^inftyint_0^{1/3}x^{2k}~mathrm dx\&stackrel?=int_0^{1/3}sum_{k=n}^infty x^{2k}~mathrm dx\&=int_0^{1/3}frac{x^{2n+1}}{1-x^2}~mathrm dx\&>int_0^{1/3}frac{x^{2n+1}}{1-(1/3)^2}~mathrm dx\&=frac1{16(n+1)9^n}end{align}



which should be sufficient. Note this also proves the slightly tighter upper bound:



$$sum_{k=n}^inftyfrac1{(2k+1)3^{2k+1}}<int_0^{1/3}frac{x^{2n+1}}{1-0^2}~mathrm dx=frac1{18(n+1)9^n}$$



and hence you can see that



$$limsup_{ntoinfty}frac{|ln2-S_{n+1}|}{|ln2-S_n|}lefrac8{81}$$






share|cite|improve this answer









$endgroup$



We can use integral remainders:



begin{align}sum_{k=n}^inftyfrac1{(2k+1)3^{2k+1}}&=sum_{k=n}^inftyint_0^{1/3}x^{2k}~mathrm dx\&stackrel?=int_0^{1/3}sum_{k=n}^infty x^{2k}~mathrm dx\&=int_0^{1/3}frac{x^{2n+1}}{1-x^2}~mathrm dx\&>int_0^{1/3}frac{x^{2n+1}}{1-(1/3)^2}~mathrm dx\&=frac1{16(n+1)9^n}end{align}



which should be sufficient. Note this also proves the slightly tighter upper bound:



$$sum_{k=n}^inftyfrac1{(2k+1)3^{2k+1}}<int_0^{1/3}frac{x^{2n+1}}{1-0^2}~mathrm dx=frac1{18(n+1)9^n}$$



and hence you can see that



$$limsup_{ntoinfty}frac{|ln2-S_{n+1}|}{|ln2-S_n|}lefrac8{81}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 12 at 1:08









Simply Beautiful ArtSimply Beautiful Art

50.7k579183




50.7k579183












  • $begingroup$
    That makes me wonder for what $alpha$ is the tail sequence equivalent to $alpha/(n9^n)$. Different question though, and something else for me try. Thanks!
    $endgroup$
    – Robert Wolfe
    Jan 12 at 1:19






  • 1




    $begingroup$
    @RobertWolfe Hint: integrate my last non-bound integral, by parts.
    $endgroup$
    – Simply Beautiful Art
    Jan 12 at 1:20




















  • $begingroup$
    That makes me wonder for what $alpha$ is the tail sequence equivalent to $alpha/(n9^n)$. Different question though, and something else for me try. Thanks!
    $endgroup$
    – Robert Wolfe
    Jan 12 at 1:19






  • 1




    $begingroup$
    @RobertWolfe Hint: integrate my last non-bound integral, by parts.
    $endgroup$
    – Simply Beautiful Art
    Jan 12 at 1:20


















$begingroup$
That makes me wonder for what $alpha$ is the tail sequence equivalent to $alpha/(n9^n)$. Different question though, and something else for me try. Thanks!
$endgroup$
– Robert Wolfe
Jan 12 at 1:19




$begingroup$
That makes me wonder for what $alpha$ is the tail sequence equivalent to $alpha/(n9^n)$. Different question though, and something else for me try. Thanks!
$endgroup$
– Robert Wolfe
Jan 12 at 1:19




1




1




$begingroup$
@RobertWolfe Hint: integrate my last non-bound integral, by parts.
$endgroup$
– Simply Beautiful Art
Jan 12 at 1:20






$begingroup$
@RobertWolfe Hint: integrate my last non-bound integral, by parts.
$endgroup$
– Simply Beautiful Art
Jan 12 at 1:20




















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