$$\text{Evaluate}\sum_{k=0}^{\infty}\bigg(\frac{1}{3k+2}-\frac{1}{3k+4}\bigg)$$ Aswini Banerjee posed this question as a challenge to me in a youtube comment. How would you solve it?

  • $\begingroup$ Would you cite the YouTube comment (for context)? $\endgroup$ – robjohn Jan 29 at 7:48
  • $\begingroup$ @robjohn, did it. Thanks. $\endgroup$ – Anubhab Ghosal Jan 29 at 8:01

$$f(x)=\sum_{k=0}^{\infty}{\bigg(\frac{x^{3k+2}}{3k+2}-\frac{x^{3k+4}}{3k+4}\bigg)}=\int_{0}^{x}{\sum_{k=0}^{\infty}\Big(t^{3k+1}-t^{3k+3}\Big)dt}=\int_{0}^{x}{\frac{t(1-t^2)}{1-t^3}dt}$$$$=x-\int_{0}^{x}{\frac{1}{t^2+t+1}dt}=x-\frac{2}{\sqrt{3}}\bigg(\arctan\frac{2x+1}{\sqrt{3}}-\frac{\pi}{6}\bigg)\implies f(1)=1-\frac{\pi}{3\sqrt{3}}$$

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    $\begingroup$ Why introduce the $x$? $\endgroup$ – Lord Shark the Unknown Jan 29 at 6:38
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    $\begingroup$ @LordSharktheUnknown, just to make the answer a bit more general, without having to do any extra steps. $\endgroup$ – Anubhab Ghosal Jan 29 at 6:40
  • $\begingroup$ Well, but how are you proving the series is uniformly convergent on some set containing the point $\large x=1$ (the necessity of proving that lies on , you are taking out the integral out of the infinite sum) ? $\endgroup$ – Bijayan Ray Jan 29 at 12:07
  • $\begingroup$ @BijayanRay, we can interchange the sum and integral for $|x| \leq 1$ by Tonelli's theorem as all the terms are positive. $\endgroup$ – Anubhab Ghosal Jan 29 at 12:27
  • $\begingroup$ Okay okay even the unform convergence follows by weierstrass m test taking the larger series 1 for $|x| \leq 1$, is not it so? $\endgroup$ – Bijayan Ray Jan 29 at 12:32

$$\sum_{k=0}^\infty\left[\frac1{3k+2}-\frac1{3k+4}\right]~=~1-\frac\pi{3\sqrt 3}$$

We may note that we can express the terms of the series slightly different as

$$\sum_{k=1}^\infty\left[\frac1{3k-1}-\frac1{3k+1}\right]=\sum_{k=1}^\infty \frac{-2}{9k^2-1}$$

The latter form can be rewritten such that we can apply a sum identity of the cotangent function. To be precise we will use the formula

$$\pi\cot(\pi z)=\frac1z+\sum_{k=1}^{\infty}\frac{2z}{z^2-k^2} $$

Thus lets rewrite the given series in the following way

$$\begin{align} \sum_{k=1}^\infty \frac{-2}{9k^2-1}&=\sum_{k=1}^{\infty}\frac{-2\frac19}{\frac1{9}-k^2}\\ &=-\frac13\left[\sum_{k=1}^{\infty}\frac{2\frac13}{\left(\frac13\right)^2-k^2}\right]\\ &=-\frac13\left(\pi\cot\left(\frac{\pi}3\right)-3\right)\\ &=1-\frac{\pi}3\cot\left(\frac{\pi}3\right) \end{align}$$

$$\therefore~\sum_{k=1}^{\infty}\frac{-2}{9k^2-1}=1-\frac{\pi}{3\sqrt 3}$$

  • $\begingroup$ Nice Solution!! $\endgroup$ – Anubhab Ghosal Jan 29 at 6:58

$$ \begin{align} \sum_{k=0}^\infty\left(\frac1{3k+2}-\frac1{3k+4}\right) &=\frac13\sum_{k=0}^\infty\left(\frac1{k+\frac23}-\frac1{k+\frac43}\right)\tag1\\ &=\frac13\sum_{k\in\mathbb{Z}}\frac1{k+\frac23}-\frac13\frac1{-1+\frac23}\tag2\\ &=\frac13\pi\cot\left(\frac{2\pi}3\right)+1\tag3\\[6pt] &=1-\frac\pi{3\sqrt3}\tag4 \end{align} $$ Explanation:
$(1)$: pull a factor of $3$ out front
$(2)$: the $k=-1$ term is not included in the original sum
$(3)$: apply $(7)$ from this answer
$(4)$: $\cot\left(\frac{2\pi}3\right)=-\frac1{\sqrt3}$

  • $\begingroup$ Nice Solution!! $\endgroup$ – Anubhab Ghosal Jan 29 at 7:36

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