# Convergence of $\sum _{n=1}^{\infty }\left\{ 1-n\log \dfrac {2n+1} {2n-1}\right\}$

I am Investigating the convergence of $$\sum _{n=1}^{\infty }\left\{ 1-n\log \dfrac {2n+1} {2n-1}\right\}$$. Now as per Cauchy's test for absolute convergence. If $\overline {\lim _{n\rightarrow \infty }}\left| u_{n}\right| ^{\dfrac {1} {n}} < 1,\sum _{n=1}^{\infty }u_{n}$ converges absolutely and obviously if $\overline {\lim _{n\rightarrow \infty }}\left| u_{n}\right| ^{\dfrac {1} {n}} > 1,\sum _{n=1}^{\infty }u_{n}$ does not converge.

I observed $$\overline {\lim _{n\rightarrow \infty }}\left| \log \left( \dfrac {2n+1} {2n-1}\right) ^{-n}\right| = \overline {\lim _{n\rightarrow \infty }}\left| \log \left( 1-\dfrac {1} {n-\dfrac {1} {2}}\right) ^{-n}\right|$$

Could i take the negative power of n outside the absolute brackets here ? I guess even if i could establish log part converges that would only show that the overall series diverges right. Is that the correct result ? Any alternative lines of attacking this problem would be much appreciated.

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I don't think you used well the convergence test. –  Beni Bogosel Mar 7 '12 at 19:12
are u saying it is not possible to take the negative power of n outside the absolute brackets here ? cause if that was allowed i'd say that the expression should be less than one as n goes to infinity. –  Hardy Mar 7 '12 at 19:20
The limit you try to calculate in your last part of the article is in fact equal to 1. And you can take the negative power of $n$ in front of the logarithm, but you can't take the negative power of $n$ out of the modulus. –  Beni Bogosel Mar 7 '12 at 19:27
Power Series of log may help. –  quartz Mar 7 '12 at 19:33
Try $$1-n \log \left( \frac{2n+1}{2n-1} \right) = 1 - n \log \left( \frac{1+\frac1{2n}}{1-\frac1{2n}} \right) = 1 - 2n \left( \frac1{2n} + \frac1{3(2n)^3} + \frac1{5(2n)^5} + \cdots \right) =\\ - \sum_{k=1,2}^{\infty} \frac1{(2k+1)(2n)^{2k}}.$$ After this you need to argue the swapping of the infinite sums and hence show it converges. –  user17762 Mar 7 '12 at 19:41

You could prove the convergence of the series using a comparison criterion. For example, calculate the limit of $|a_n|/(\frac{1}{n^2})$. You should then calculate $$\lim_{n \to \infty}\left| n^2-n^3\log \left(\frac{2n+1}{2n-1} \right)\right|$$.

For this calculation, the simplest method I could think of was expanding in Taylor series.

$$\log(x+1)-\log(x-1)=\log \left(1+\frac{1}{x}\right)-\log\left(1-\frac{1}{x}\right)=2\sum_{k \text{ odd}}\frac{1}{kx^k}$$

Then you have to calculate

$$\lim_{n \to \infty} \left|n^2-n^3\cdot 2\left(\frac{1}{2n}+\frac{1}{3(2n)^3}+\frac{1}{5(2n)^5}+... \right)\right|=\frac{1}{12}$$

Therefore $\sum a_n$ is absolutely convergent, and in particular convergent.

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Nicely done, though i did n't follow how u got 1/12 as the result –  Hardy Mar 7 '12 at 19:57
Also as per comparison series do n't we need to find a series which forms and upper bound to our series and show since that converges absolutely ours would too, in the answer you posted i am unsure what series that would be as we only seem to change our series and operate on that. –  Hardy Mar 7 '12 at 20:04
There are three comparison criteria. Two with inequalities, and the last one, says that $a_n /b_n$ tends to a positive, bounded limit, then the series $a_n, b_n$ have the same nature. The comparison criteria with inequalities are obviously equivalent to the one with the limit. –  Beni Bogosel Mar 7 '12 at 21:24
As for the limit in my answer, try and open the bracket, and see which terms remain. It's quite easy. –  Beni Bogosel Mar 7 '12 at 21:25
+1. Nicely done. This also quantifies how the series converges like the series $\sum \frac1{n^2}$. –  user17762 Mar 7 '12 at 22:15

$1-n\log\left(\frac{2n+1}{2n-1}\right)= 1-n\log\left(1+\frac{1}{n-\frac{1}{2}}\right)\sim 1-\frac{n}{n-\frac{1}{2}}+\frac{n}{2\left(n-\frac{1}{2}\right)^{2}}=\frac{1}{(2n-1)^{2}}\sim \frac{1}{4 n^{2}}$ then the series converges.

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HINT: Relate $\lim_{n \to \infty}\left| n^2-n^3\log \left(\frac{2n+1}{2n-1} \right)\right|$ to $\lim_{n \to \infty} n-n^2\log \left(\frac{n+1}{n} \right)$whose limit is $\frac{1}{2}$ and can be almost elementarily proved by replacing n by $\frac{1}{x}$ when $x \rightarrow 0$. The second limit comes from a highschool book, 11th grade, often met during courses.

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This question has an interesting aspect that merits being commented, which is that we can compute a closed form for the sum using Mellin transforms and harmonic sums.

Introduce the sum $S(x)$ given by $$S(x) = \sum_{n\ge 1} \left(1- xn \log\frac{2xn+1}{2xn-1}\right)$$ so that we are interested in $S(1).$

As mentioned before the sum term is harmonic and may be evaluated by inverting its Mellin transform.

Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have $$\lambda_k = 1, \quad \mu_k = k \quad \text{and} \quad g(x) = 1-x\log\frac{2x+1}{2x-1} = 1-x\log\left(1+\frac{2}{2x-1}\right).$$

The abscissa of convergence of $$\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \zeta(s)$$ is $\Re(s)>1.$

Now to find the Mellin transform $g^*(s)$ of $g(x)$ the fundamental strip being $\langle 0,2\rangle$ (which contains the abscissa of convergence of the Dirichlet series sum term) we first use integration by parts to get $$\int_0^\infty \left(-x\log\left(1+\frac{2}{2x-1}\right)\right) x^{s-1} dx \\= \left[\left(-x\log\left(1+\frac{2}{2x-1}\right)\right) \frac{x^{s+1}}{s+1}\right]_0^\infty - \int_0^\infty \frac{4}{4x^2-1} \frac{x^{s+1}}{s+1} dx = - \int_0^\infty \frac{4}{4x^2-1} \frac{x^{s+1}}{s+1} dx.$$ with the fundamental strip being $\langle -1, 0\rangle$. It becomes evident that we require the following Mellin transform: $$h^*(s) = \int_0^\infty h(x) x^{s-1} dx$$ where $$h(x) = \frac{4}{4x^2-1}.$$ This transform integral is not strictly speaking convergent but we can compute its principal value by using a semicircular contour in the upper half plane that is traversed clockwise and picks up half the residues at the two poles at $\pm 1/2.$ This gives $$h^*(s) (1-e^{i\pi s}) = \frac{1}{2} \times 2 \pi i \left(\mathrm{Res}(h(x) x^{s-1}; x=1/2) + \mathrm{Res}(h(x) x^{s-1}; x=-1/2)\right)$$ which gives $$h^*(s) (1-e^{i\pi s}) = \frac{1}{2} \times 2 \pi i ((1/2)^{s-1}-(-1/2)^{s-1}) = 2^{-s} \times 2 \pi i (1+(-1)^s).$$ This yields $$h^*(s) = 2^{-s} \times 2 \pi i \frac{1+e^{i\pi s}}{1-e^{i\pi s}} = 2^{-s} \times 2 \pi i \frac{e^{-i\pi s/2}+e^{i\pi s/2}}{e^{-i\pi s/2}-e^{i\pi s/2}} = -\frac{2}{2^s} \pi \cot(\pi s/2).$$ Returning to $g^*(s)$ we have $$g^*(s) = \frac{2}{2^{s+2}} \frac{1}{s+1} \pi \cot(\pi (s+2)/2) = \frac{1}{2^{s+1}} \frac{1}{s+1} \pi \cot(\pi s / 2).$$

Therefore the Mellin transform $Q(s)$ of $S(x)$ is given by $$\frac{1}{2^{s+1}} \frac{1}{s+1} \pi \cot(\pi s / 2)\zeta(s).$$

The Mellin inversion integral here is $$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$ which we evaluate by shifting it to the right for an expansion about infinity.

We must sum the residues at the poles at the positive even integers of $Q(s)/x^s$ which are $$\mathrm{Res}(Q(s)/x^s; s=2q) = \frac{1}{2^{2q+1}} \frac{1}{2q+1} \times 2 \times \zeta(2q) \frac{1}{x^{2q}} \\ =\frac{1}{2^{2q+1}} \frac{1}{2q+1} \times 2 \times \frac{(-1)^{q+1} B_{2q} (2\pi)^{2q}}{2\times (2q)!} \frac{1}{x^{2q}} = \frac{1}{2} \times \frac{(-1)^{q+1} B_{2q} \pi^{2q}}{(2q+1)!} \frac{1}{x^{2q}}.$$

We thus require the sum $$- \frac{1}{2} \sum_{q\ge 1} \frac{i^{2q} B_{2q} \pi^{2q}}{(2q+1)!} \frac{1}{x^{2q}}.$$

The exponential generating function of the even Bernoulli numbers is $$-1 + \frac{1}{2} t + \frac{t}{e^t-1} = \sum_{q\ge 1} B_{2q} \frac{t^{2q}}{(2q)!}.$$ Integrate this to obtain $$-t - \frac{t^2}{4} + t\log(1-e^t) + \mathrm{Li}_2(e^t).$$ The constant that appeared during the integration was the well-known zeta function value $\mathrm{Li}_2(1) = \zeta(2) = \pi^2/6$ so that we finally have $$\sum_{q\ge 1} B_{2q} \frac{t^{2q}}{(2q+1)!} = \frac{1}{t} \left(-\frac{\pi^2}{6}-t - \frac{t^2}{4} + t\log(1-e^t) + \mathrm{Li}_2(e^t)\right).$$ To conclude use another well known zeta function value which is $\mathrm{Li}_2(-1) = -\pi^2/12$ (derived from $\mathrm{Li}_2(1)$) and put $t=i\pi /x = i\pi$ to obtain that (we lose the minus sign because we are shifting to the right) $$S(1) = \frac{1}{2} \frac{1}{i\pi} \left(-\frac{\pi^2}{6}-i\pi + \frac{\pi^2}{4} + i\pi\log 2 + \mathrm{Li}_2(-1)\right) \\ = \frac{1}{2} \frac{1}{i\pi} \left(-i\pi + i\pi \log 2 \right) = \frac{1}{2} (\log 2 - 1).$$

Remark. The integral of the generating function of the Bernoulli numbers is easily verified by differentiation: $$\left(-\frac{t^2}{2} + t\log(1-e^t) + \mathrm{Li}_2(e^t)\right)' = -t + \frac{t}{1-e^t} (-e^t) + \log(1-e^t) + \left(\sum_{n\ge 1} \frac{e^{nt}}{n^2}\right)' \\ = \frac{-t+te^t}{1-e^t} - \frac{te^t}{1-e^t} + \log(1-e^t) + \sum_{n\ge 1} \frac{e^{tn}}{n} \\ = \frac{-t}{1-e^t} + \log(1-e^t) - \log(1-e^t) = \frac{t}{e^t-1}.$$

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