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

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 $$\limsup_{n\rightarrow\infty}\frac{|\ln 2-S_{n+1}|}{|\ln 2-S_{n}|}<1$$ where $\{S_n\}$ are the partial sums?

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

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

We can use integral remainders:

\begin{align}\sum_{k=n}^\infty\frac1{(2k+1)3^{2k+1}}&=\sum_{k=n}^\infty\int_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:

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

and hence you can see that


  • $\begingroup$ 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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