Has this infinite sum $\sum _{i=1}^{\infty } p^i \log (b i+a)$ any known solution?

Question

I am wondering if exist a known solution for this kind of infinite sum

$$ \sum _{i=1}^{\infty } p^i \log (a i + b) $$ where $p,a,b$ are real and $p\leq 1$. ...or even an approximation of the exact solution.

@Alex p is a variable just like a,b, and i. Op doesn't need to elaborate. — Zach466920, Apr 10 '15 at 19:56
I think for $|p|<1$ it should converge (comparing to integral $\int_{1}^{\infty} t^x \log x dx$) — Alex, Apr 10 '15 at 20:03

score 2 · Accepted Answer · answered Apr 10 '15 at 20:06

2

Assume $|p|<1$. One may recall the Lerch transcendent given by $$ \Phi(p, s, \alpha) = \sum_{i=0}^\infty \frac { p^i} {(i+\alpha)^s}. $$ Then observing that $$ \partial_s\left.\frac {1} {(i+\alpha)^s}\right|_{s=0}=-\left.\frac {\log (i+\alpha)} {(i+\alpha)^s}\right|_{s=0}=-\log (i+\alpha) $$ leads readily to

$$ \sum _{i=1}^{\infty } p^i \log (a i + b)=-p \: \left.\partial_s\Phi\! \left(p,s,1+\frac{a}{b}\right)\right|_{s=0}+\frac{p}{1-p}\:\log b. $$

answered Apr 10 '15 at 20:06

Olivier Oloa

120,989

Thanx Olivier. Has $$ \partial_s \Phi (p,s,1+\frac{a}{b}) $$ any aproximate form? – emanuele Apr 10 '15 at 20:23
1

@emanuele Yes, by integral representation of $\Phi$, you may obtain numerical approximation. Thanks. – Olivier Oloa Apr 10 '15 at 20:28

Zach466920 · Answer 2 · 2015-04-10T20:41:26.287

0

According to wolfram your series is quite complicated. You get this... $${{\log(a)} \over {1-p}}-p*\Phi(p,0,b/a)$$ where the second part is the Lerch Transcendental(I've been informed this needs to be differentiated). It looks like a solution, but math has a funny way with equality. Eventually doing everything out, you just realize this function is just a circular definition.

edited Apr 10 '15 at 20:41

answered Apr 10 '15 at 20:06

Zach466920

8,341

You should differentiate $\displaystyle \Phi(p, s, \alpha)$. – Olivier Oloa Apr 10 '15 at 20:10
@OlivierOloa This is totally out of my depth. Why differentiate? But what do I know, if you ask me, wolfram could very well be wrong on this one. All I know for sure is that this solution is not as useful as it appears. – Zach466920 Apr 10 '15 at 20:12
"Why differentiate?" In order to get the factor $\log(ai+b)$ in the summand. – Olivier Oloa Apr 10 '15 at 20:16
@OlivierOloa Oh, Ok that makes since. – Zach466920 Apr 10 '15 at 20:28
wolfram show the partial derivative too on the lerch function. you should change your answer. – emanuele Apr 10 '15 at 20:30
@emanuele nope not anything I could see, I was missing a p though. – Zach466920 Apr 10 '15 at 20:41
which version of mathematica you have? – emanuele Apr 10 '15 at 21:41

score 0 · Answer 3 · answered Apr 10 '15 at 20:48

0

A special case would be $$ \sum_{n=1}^\infty \frac{\log n}{2^n} \approx 0.507833922868438392 , $$ which is unknown to the Inverse Symbolic Calculator

answered Apr 10 '15 at 20:48

GEdgar

111,679

Has this infinite sum $\sum _{i=1}^{\infty } p^i \log (b i+a)$ any known solution?

3 Answers3