2

I have many sums to compute but all of them are similars to each other. There are of this kind: $$\sum_{n=1}^{\infty} \frac{n^a}{b^n}$$ $$a,b \in \mathbb{Z}$$ $$a \in [1,\infty)$$ $$b \in (a,\infty)$$

For instance: $$\sum_{n=1}^{\infty} \frac{n^3}{5^n}$$

openspace
  • 6,470
Piotr Wasilewicz
  • 306
  • In general you could use this link: https://en.wikipedia.org/wiki/Lerch_zeta_function But for the specific sum you mentioned things are simpler ... – Math-fun Jan 09 '17 at 12:36
  • If I were merely trying to prove convergence, I would appeal to a ratio test. Starting from scratch I'd be tempted to think about these as power series in $1/b$. – hardmath Jan 09 '17 at 12:40
  • 1
    See here: http://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-infty-n1xn (with $x=1/b$). (That question is about $a=1$, but there are also answers which explain how to do it for any $a$.) – Hans Lundmark Jan 09 '17 at 12:59

1 Answers1

1

Let $f_1(b)=\sum_{n=1}^{\infty}nb^{-n}.$ Then $$f_1(b)/b=\sum_{n=1}^{\infty}nb^{-n-1}=-\sum_{n=1}^{\infty}\frac {db^{-n}}{db}=$$ $$=-\frac {d}{db} \sum_{n=1}^{\infty} b^{-n}=-\frac {d(b-1)^{-1}}{db}=\frac {1}{(b-1)^2 }.$$ So $f_1(b)=\frac {b}{(b-1)^2}.$

Let $f_2(b)=\sum_{n=1}^{\infty}n^2b^{-n}.$ Then $$f_2(b)/b= \sum_{n=1}^{\infty}n^2b^{-n-1}=-\frac {d}{db}\sum_{n=1}^{\infty}nb^{-n}=-\frac {df_1(b)}{db}.$$ So $f_2(b)=-bf'_1(b).$

Et Cetera.

DanielWainfleet
  • 57,985
  • Thanks. I think it is not the simplest way (I have big "a" and "b"), but maybe I will find some recursive formula after some computions. – Piotr Wasilewicz Jan 24 '17 at 09:48