# Summing an alternating series of type $(-1)^{k-1} \frac1{1 + ak^{-1}}$

I have the following alternating series:

$$\sum_{k=1}^{+\infty} (-1)^{k-1} \dfrac{1}{1 + \alpha{}k^{-1}}$$

with $\alpha<1$, and I'm trying to evaluate the sum. This series arose from a previous question I posted yesterday. Without citing too much from that post, the original expression is bounded, so I think this sum should also be bounded. It seems likely, due to the small coefficient and the inverse power.

I have looked around for some hints on how to proceed, but I didn't find anything useful.

Anything I can use to evaluate this?

Thanks,

• $\frac{1}{1 + \alpha{}k^{-1}}\to 1$ as $k\to \infty$ – Robert Z May 24 '18 at 9:51
• @RobertZ and this implies that it will not converge? – KaiserHaz May 24 '18 at 10:21
• – Robert Z May 24 '18 at 10:22
• I see your point. Thank you, – KaiserHaz May 24 '18 at 10:29

As stated by @RobertZ, this sum does not converge. However, one can somehow assign a principal value for it.

Note that $$\frac1{1+ak^{-1}}=1-\frac{a}{k+a}$$

If one accepts that $$\sum^\infty_{k=1} (-1)^k=0$$ it is easy to see your sum equals $$\color{RED}{a\cdot\Phi(-1,1,a)-1}$$ where $\Phi$ is the Lerch transcendent.

$$\Phi(z,s,q)=\sum^\infty_{k=0}\frac{z^k}{(k+q)^s}$$

• Say if the condition for sum of alternating ones is instead not equal to zero (maybe $1$ or $-1$, that is, if the infinity is a 'relatively large even or odd number'), then would there be any difference in the resulting expression? – KaiserHaz May 24 '18 at 10:37
• @KaiserHaz Then of course $\sum(-1)^k$ has a definite value. However, the sum making the Lerch transcendent should remain the same. – Szeto May 24 '18 at 10:43
• How does that change the result? Would it be multiplied by $1$ or $-1$? – KaiserHaz May 24 '18 at 11:09
• @KaiserHaz The result will be added or subtracted by one. – Szeto May 24 '18 at 11:18

I estimated lower and upper limits of the sum:

$S=\sum\limits_{k=1}^{+\infty} (-1)^{k-1} \dfrac{1}{1 + \alpha{}k^{-1}}$

Separate the sum according to the odd and even value of $k$s then

$S=\sum\limits_{i=0}^{+\infty}\big( (-1)^{2i} \dfrac{1}{1 + \frac{\alpha}{2i+1}}-(-1)^{2i+1} \dfrac{1}{1 + \frac{\alpha}{2i+2}}\big)$ performing the possible simplifications I got:

$S=\sum\limits_{i=0}^{+\infty}\dfrac{-\alpha}{(2i+2+\alpha)(2i+1+\alpha)}$ then -S was forced to the following limits:

$\sum\limits_{i=0}^{+\infty}\dfrac{\alpha}{(2i+2+\alpha)^2}\lt -S\lt \sum\limits_{i=0}^{+\infty}\dfrac{\alpha}{(2i+1+\alpha)^2}$

$\frac{\alpha}{4}\sum\limits_{i=0}^{+\infty}\dfrac{1}{(i+\frac{(2+\alpha)}{2})^2}\lt -S\lt \frac{\alpha}{4}\sum\limits_{i=0}^{+\infty}\dfrac{1}{(i+\frac{(1+\alpha)}{2})^2}$

Finally $\frac{\alpha}{4}\zeta(2,\frac{(2+\alpha)}{2})\lt-S \lt \frac{\alpha}{4}\zeta(2,\frac{(1+\alpha)}{2})$ where $\zeta(s,q)$ is the Hurwitz zeta function.