4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
9
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Why introduce the $x$? $\endgroup$ – Lord Shark the Unknown Jan 29 at 6:38
  • 1
    $\begingroup$ @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
3
$\begingroup$

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

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

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

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

$\endgroup$
  • $\begingroup$ Nice Solution!! $\endgroup$ – Anubhab Ghosal Jan 29 at 6:58
2
$\begingroup$

$$ \begin{align} \sum_{k=0}^\infty\left(\frac1{3k+2}-\frac1{3k+4}\right) &=\frac13\sum_{k=0}^\infty\left(\frac1{k+\frac23}-\frac1{k+\frac43}\right)\tag1\\ &=\frac13\sum_{k\in\mathbb{Z}}\frac1{k+\frac23}-\frac13\frac1{-1+\frac23}\tag2\\ &=\frac13\pi\cot\left(\frac{2\pi}3\right)+1\tag3\\[6pt] &=1-\frac\pi{3\sqrt3}\tag4 \end{align} $$ 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
$(4)$: $\cot\left(\frac{2\pi}3\right)=-\frac1{\sqrt3}$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.