MHB Efficient Methods for Evaluating Complex Sums: A Scientific Approach

[Saitama]

While doing an another problem, I came across the following sum and I have no idea about how one should go about evaluating it.

$$\sum_{k=0}^{\infty} (-1)^k\left(\frac{1}{(3k+2)^2}-\frac{1}{(3k+1)^2}\right)$$

Wolfram Alpha gives $-\frac{2\pi^2}{27}$ as the result but I have absolutely no idea how it got this. In alternate forms for this solution, it shows digamma and generalised Zeta function but I would like a solution without using these. You may use $\zeta(2)=\pi^2/6$ because I think it would necessary for this sum.

Any help is appreciated. Thanks!

 

[DreamWeaver]

Hi Pranav! :D

Ostensibly, the Hurwitz Zeta functions - of which the Riemann Zeta is but a special case - are one and the same as the Polygamma functions, so even without 'appearing to use them' in a solution, you will do anyway, indirectly.

I'll sketch out a proof, but it'll take a little while for me to type the code. Back in a bit... ;)

 

[DreamWeaver]

NOTE: I don't quite get the same answer, but it's out by just a small fraction. I can't find any mistakes here, but that certainly doesn't rule it out... (Headbang) Nonetheless, I hope the methodology will at least be useful! lol
Split the series into two:

$$\sum_{k=0}^{\infty}\frac{(-1)^k}{(3k+2)^2} - \sum_{k=0}^{\infty}\frac{(-1)^k}{(3k+1)^2}$$

And then split each of those into two, by splitting the index k into odd and even terms 2k and 2k+1, thus eliminating the alternating signs:$$\sum_{k=0}^{\infty}\frac{1}{(6k+2)^2} - \sum_{k=0}^{\infty}\frac{1}{(6k+5)^2} -
\sum_{k=0}^{\infty}\frac{1}{(6k+1)^2} + \sum_{k=0}^{\infty}\frac{1}{(6k+4)^2}=$$
$$-\sum_{k=0}^{\infty} \left[ \frac{1}{(6k+1)^2} + \frac{1}{(6k+2)^2} + \frac{1}{(6k+3)^2} + \frac{1}{(6k+4)^2} + \frac{1}{(6k+5)^2 } + \frac{1}{(6k+6)^2}
\right] +$$

$$2\, \sum_{k=0}^{\infty}\frac{1}{(6k+2)^2} +
\sum_{k=0}^{\infty}\frac{1}{(6k+3)^2} +
2\, \sum_{k=0}^{\infty}\frac{1}{(6k+4)^2} +
\sum_{k=0}^{\infty}\frac{1}{(6k+6)^2}=$$
$$-\sum_{k=1}^{\infty}\frac{1}{k^2} + 2\, \sum_{k=0}^{\infty}\frac{1}{(6k+2)^2} +
\sum_{k=0}^{\infty}\frac{1}{(6k+3)^2} +
2\, \sum_{k=0}^{\infty}\frac{1}{(6k+4)^2} +
\sum_{k=0}^{\infty}\frac{1}{(6k+6)^2}=$$
$$-\zeta(2) + 2\, \sum_{k=0}^{\infty}\frac{1}{(6k+2)^2} +
\sum_{k=0}^{\infty}\frac{1}{(6k+3)^2} +
2\, \sum_{k=0}^{\infty}\frac{1}{(6k+4)^2} +
\sum_{k=0}^{\infty}\frac{1}{(6k+6)^2}=$$
$$-\zeta(2) +2\, \sum_{k=0}^{\infty} \left[ \frac{1}{(6k+2)^2} + \frac{1}{(6k+4)^2}\right] +
\sum_{k=0}^{\infty} \left[ \frac{1}{(6k+3)^2} + \frac{1}{(6k+6)^2}\right] = $$
$$-\zeta(2) +\frac{2}{2^2}\, \sum_{k=0}^{\infty} \left[ \frac{1}{(3k+1)^2} + \frac{1}{(3k+2)^2}\right] +
\frac{1}{3^2}\, \sum_{k=0}^{\infty} \left[ \frac{1}{(3k+1)^2} + \frac{1}{(3k+2)^2}\right] =$$
$$-\zeta(2) +\frac{11}{18}\, \sum_{k=0}^{\infty} \left[ \frac{1}{(3k+1)^2} + \frac{1}{(3k+2)^2}\right] = $$
$$-\zeta(2) +\frac{11}{18}\, \sum_{k=0}^{\infty} \left[ \frac{1}{(3k+1)^2} + \frac{1}{(3k+2)^2} + \frac{1}{(3k+3)^2} \right] - \frac{11}{18}\, \sum_{k=0}^{\infty} \frac{1}{(3k+3)^2} = $$
$$-\zeta(2) +\frac{11}{18}\, \sum_{k=1}^{\infty}\frac{1}{k^2} - \frac{11}{18}\, \frac{1}{3^2}\, \sum_{k=0}^{\infty}\frac{1}{(k+1)^2} =$$$$-\zeta(2) + \frac{11}{18}\, \zeta(2) -\frac{11}{162}\, \zeta(2) = -\frac{74}{162}\, \zeta(2) = $$$$-\frac{74}{162}\, \frac{\pi^2}{6} = -\frac{37\pi^2}{486}
$$
NOTE:

$$\frac{2}{27} = 0.0740740740...$$

$$\frac{37}{486} = 0.076131687...$$Looks like I need help on this one too! (Sun)

 

[alyafey22]

I' m sorry I only have access to the internet via my phone and writing latex using my phone will be very difficult so here is a hint

k can be written as 3k , 3k+1, 3k+2

Note that you have the last two terms in your sum.

 

[DreamWeaver]

I' m sorry I only have access to the internet via my phone and writing latex using my phone will be very difficult so here is a hint

k can be written as 3k , 3k+1, 3k+2

Note that you have the last two terms in your sum.

Hello, friend! (Hug)

That would have been my instinctive method of attack too, but then it leads to an evaluation in terms of Digammas (or equivalently, Dirichlet L-series), which Pranav was hoping to avoid.

ps. Hope the phone/internet thing is just a temporary situation...

 

[alyafey22]

Hey DW

Start by the alternating zeta function

$$\sum \frac{(-1)^k}{k^2} $$

Then separate it into three terms.

Ps: I hope so but that might take a long time.

 

[DreamWeaver]

Hey DW

Start by the alternating zeta function

$$\sum \frac{(-1)^k}{k^2} $$

Then separate it into three terms.

Ps: I hope so but that might take a long time.

I see what you mean there, but I'm not sure that will work; that's why I instinctively split it into 6 parts/series (4 to begin with, then two more added), since otherwise you end up with (new) signed coefficients $$(-1)^{3k}\, $$, $$(-1)^{3k+1}\, $$, and $$(-1)^{3k+2}\,$$ , all of which, themselves, alternate in sign.

Unless I'm misunderstanding exactly how you mean applying the split...?

[Sorry]

 

[alyafey22]

$$\sum \frac{(-1)^k}{k^2}=\sum \frac{(-1)^k}{9k^2}-\sum \frac{(-1)^k}{(3k+1)^2}+\sum \frac{(-1)^k}{(3k+2)^2}$$

$$\sum \frac{(-1)^k}{(3k+2)^2}-\frac{(-1)^k}{(3k+1)^2}=\left(1- 1/9\right) \sum \frac{(-1)^k}{k^2}=\frac{-2\pi^2}{27}$$

 

[DreamWeaver]

$$\sum \frac{(-1)^k}{k^2}=\sum \frac{(-1)^k}{9k^2}-\sum \frac{(-1)^k}{(3k+1)^2}+\sum \frac{(-1)^k}{(3k+2)^2}$$

$$\sum \frac{(-1)^k}{(3k+2)^2}-\frac{(-1)^k}{(3k+1)^2}=\left(1- 1/9\right) \sum \frac{(-1)^k}{k^2}=\frac{-2\pi^2}{27}$$

Brilliant! I bow (seriously) before the master... (Sun)

Thanks for that, Zaid! (Hug)ps. Sorry for sort-of high-jacking your thread, Pranav!

 

[Saitama]

$$\sum \frac{(-1)^k}{k^2}=\sum \frac{(-1)^k}{9k^2}-\sum \frac{(-1)^k}{(3k+1)^2}+\sum \frac{(-1)^k}{(3k+2)^2}$$

$$\sum \frac{(-1)^k}{(3k+2)^2}-\frac{(-1)^k}{(3k+1)^2}=\left(1- 1/9\right) \sum \frac{(-1)^k}{k^2}=\frac{-2\pi^2}{27}$$

Great! Thanks a lot! (Sun)

Thank you Dreamweaver for trying to help me out with the problem. :)