Seeking for $\alpha\sum_{k=1}^{\infty}{(-1)^{k-1}\over k(e^{k\pi}-1)}+\beta\sum_{k=1}^{\infty}{(-1)^{k-1}\over k(e^{k\pi}+1)}=f(\alpha,\beta)$

Question

We are seeking the closed form of $f(\alpha,\beta)$

Let

$$P=\sum_{k=1}^{\infty}{(-1)^{k-1}\over k(e^{k\pi}-1)}$$ and

$$Q=\sum_{k=1}^{\infty}{(-1)^{k-1}\over k(e^{k\pi}+1)}$$

We have a closed form of

$$\ln(2)=12P+4Q$$

Let $$\alpha P+\beta Q=f(\alpha,\beta)$$

We are assuming that $\alpha$ and $\beta$ are integers

We are wondering if there is a closed form for $f(\alpha,\beta)?$

Have a look at I.J.Zucker - THE SUMMATION OF SERIES OF HYPERBOLIC FUNCTIONS, 1979. — Jack D'Aurizio, Jun 20 '17 at 19:18
You may have a look at this answer https://math.stackexchange.com/a/1912949/72031 where it is established that $\log 2=12P+4Q$. You can get values of $P, Q$ in closed form using this technique. — Paramanand Singh, Jun 21 '17 at 05:40

orlp · Accepted Answer · 2017-06-20T22:41:37.783

0

$$P + Q = \lim_{n\to\infty}\left ( \sum_{k=1}^{n}{(-1)^{k-1}\over k(e^{k\pi}-1)} + \sum_{k=1}^{n}{(-1)^{k-1}\over k(e^{k\pi}+1)}\right )$$

$$P + Q = \lim_{n\to\infty}\left ( \sum_{k=1}^{n}{(-1)^{k-1}\over k} \left({1 \over e^{k\pi}-1} + {1 \over e^{k\pi}+1} \right )\right )$$

$$P + Q = \lim_{n\to\infty}\left ( \sum_{k=1}^{n}{(-1)^{k-1}\over k} \left({2e^{\pi k} \over (e^{k\pi}-1)(e^{k\pi}+1)} \right )\right )$$

$$P + Q = \lim_{n\to\infty}\left ( \sum_{k=1}^{n}{(-1)^{k-1}\over k} \text{csch}(\pi k)\right )$$

$$P + Q = \sum_{k=1}^\infty (-1)^{k+1}\frac{ \text{csch}(\pi k)}{k} = \frac{1}{12}(\ln(64) - \pi)$$

The final identity was derived using Wolfram Alpha, but was a mystery to me. I asked a new question to demystify it, which Jack D'Aurizio answered.

We have $P + Q = \frac{1}{12}(\ln(64) - \pi)$ and $12P + 4Q = \ln(2)$, giving us:

$$P = -\frac{1}{24}(\ln(8) - \pi), \quad Q = \frac{1}{8}(\ln(32) - \pi)$$

edited Jun 20 '17 at 22:41

answered Jun 20 '17 at 20:43

orlp

@orip please explain how you got $P+Q = \frac{1}{2}(\ln(64) - \pi)$ from the Sum? – unseen_rider Jun 20 '17 at 20:54
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@unseen_rider I added the derivation of $\text{csch}$, sadly the most interesting part was found with Wolfram Alpha - I'm still searching a reference for the identity. – orlp Jun 20 '17 at 21:10
@unseen_rider I asked a new question for the identity, which Jack D'Aurizio answered. – orlp Jun 20 '17 at 22:19
That should be $\frac{1}{12}(\log{(64)} - \pi)$ here and in the other question. – sharding4 Jun 20 '17 at 22:25
@sharding4 Oops, fixed. – orlp Jun 20 '17 at 22:43
@NightOfPower Excuse me? – orlp Jun 23 '17 at 10:29

1 Answers1