# Evaluating $\sum\limits_{n=1}^{\infty} \frac1{n}\left(\frac{np}{p+n}\right)^{n+1}$ for $0<p<1$

I'm trying to find the value of the following sum

$$\sum_{n=1}^{\infty} \frac1{n}\left(\frac{np}{p+n}\right)^{n+1}$$

where $0<p<1$. Any ideas? Thanks.

-
My guess is that there is no nice closed form solution. – Eric Naslund Jun 2 '11 at 1:24
Certainly the $n^{n+1}$ factor will likely preclude an expression in terms of currently known functions... – J. M. Jun 2 '11 at 4:30
Guessing you can't guess is a funny guess... the only thing I can say is that I am sure it is absolutely convergent (which I believe you noticed since you worked on it... but I was suspicious at first so I'll just make the worried people stop wondering ) :$$\sum_{n=1}^{\infty} \frac 1n \left( \frac {np}{p+n} \right)^{n+1} = \sum_{n=1}^{\infty} \frac 1n \left( \frac {p}{\frac pn + 1} \right)^{n+1} \le \sum_{n=1}^{\infty} \frac 1n p^{n+1} \le \sum_{n=1}^{\infty} p^{n+1} = \frac 1{1-p}$$ because $p/n+1 \ge 1$, so $1 \ge \frac 1{p/n+1}$. Now I can sleep. Good luck with it though. – Patrick Da Silva Jun 2 '11 at 5:18
Reordering summation/collecting like powers of p I get the following expression $$f(p) = p \sum_{j=1}^{\infty} (-1)^j c_j p^j$$ where $$c_j = \sum_{k=1}^j (-1)^k \binom {j}{k} k^{k-j-1}$$. The $c_j$ approximate $-j$ so the $\delta$ in my previous comment is now $\delta_j = c_j +j$ (I hope, all indexes and signs are correctly taken from my sketchpad...) – Gottfried Helms Jun 2 '11 at 11:05
Not to be pedantic, but this is not a "sum of a series", but just "a series" (=' sum of the terms of a sequence'). – leonbloy Jul 10 '11 at 14:35

If $0<p<1$ then $\dfrac{np}{n+p}<\dfrac{np}{n}=p$. Thus $$\sum_{n=1}^{\infty } \frac{1}{n}\left(\frac{np}{p+n}\right)^{n+1}< \sum_{n=1}^{\infty } \frac{1}{n}\left(p^{n+1}\right) = . . .(1) . . .= -p \ln(1-p)$$ (1): If $0<p<1$, then $$\sum_{n=1}^{\infty }\frac{1}{n}\left(p^{n+1}\right) = p \sum_{n=1}^{\infty } \frac{1}{n}\left(p^{n}\right) =p\int \sum_{n=1}^{\infty} \ p^{n-1} dp =p\int \frac{1}{1-p} dp = -p\ln(1-p)$$

-
Your point being? – J. M. Sep 24 '11 at 9:02
I don't know why the site can not loaded completely.If you go to edit page ,you can see the complete my answer. Thanks – saeed sani Sep 24 '11 at 14:55
I checked the edit, what's displayed to me is exactly what's in the edit history. I am not sure the problem is with loading... – Srivatsan Sep 24 '11 at 14:57
it's good. my answer result a Upper bound for summation that it is very near real answer. – saeed sani Sep 24 '11 at 15:09
@saeed, I formatted your post (please see the edits in the source, you will see that life with LaTeX can be much simpler than what you make it). – Did Sep 24 '11 at 16:55

I can't comment so that's why i'm giving an answer but isn't it true that $\frac 1n \left( \frac {np}{p+n} \right)^{n+1} = \frac 1n \frac {n^{n+1}p^{n+1}}{(p+n)^{n+1}} = \frac {n^np^{n+1}}{(p+n)^{n+1}} = \left( \frac {n^{-1}p}{p+n} \right)^{n+1}$ so couldn't you just sum $\sum _{n=1}^{\infty } \left( \frac {n^{-1}p}{p+n} \right)^{n+1}$ using $\frac {a}{1-r}$ with $a = \left( \frac {p}{p+n} \right)^2$ and $r = \frac {n^{-1}p}{p+n}$ which yields $\sum _{n=1}^{\infty } \left( \frac {n^{-1}p}{p+n} \right)^{n+1} = \frac {\left( \frac {p}{p+n} \right)^2}{1-\frac {n^{-1}p}{p+n}} = \frac{p^2}{(p+n)^2} \frac {p+n}{p+n- n^{-1}p}$? or have i missed something? hope that helps.

-
Your first equation is not correct, because $$(n^{-1})^{n+1}\neq n^n.$$ Instead, $$(n^{-1})^{n+1}=n^{-(n+1)}=n^{-n-1}.$$ – Zev Chonoles Jul 10 '11 at 15:10
Also, the geometric series formula only works if $r$ is fixed, in your case $r = \frac {n^{-1}p}{p+n}$ depends on $n$. – N. S. Jul 10 '11 at 16:10
I thought there was something wrong, probably shouldn't do maths that late. Sorry guys. – Ross Pure Jul 11 '11 at 6:01
Also, I can take down the answer if you want. – Ross Pure Jul 11 '11 at 6:03
Don't do that . – Patrick Da Silva Oct 1 '11 at 18:11