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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.

J. M. ain't a mathematician
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a06e
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    My guess is that there is no nice closed form solution. – Eric Naslund Jun 02 '11 at 01:24
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    Certainly the $n^{n+1}$ factor will likely preclude an expression in terms of currently known functions... – J. M. ain't a mathematician Jun 02 '11 at 04:30
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    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 02 '11 at 05:18
  • If you define this as function $f(p)$ then the terms of the powerseries begin with $ p^2 - 3/2p^3 + 31/12p^4 - 31/9p^5 + 10007/2160p^6 - ...$ Let's write x for p now... Roughly this goes to $f(x) = x*\sum_{k=1}^{\infty} (-1)^k (k-\delta_k) x^k $ where the sequence $delta_k$ looks similar to the harmonic series and doesn't sum up nicely to some estimate of bound. (just another hint, that it should be convergent for $|x|< 1$ ) – Gottfried Helms Jun 02 '11 at 10:04
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    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 02 '11 at 11:05
  • Thank you all for your insights. Something like this is what I expected. I'm not sure if a closed form can be found, so brainstorm is just the way to go now. – a06e Jun 03 '11 at 13:43
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    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
  • You should somehow use that $$\Bigl({1\over 1+{p\over n}}\Bigr)^n\doteq e^{-p}\ .$$ – Christian Blatter Jul 10 '11 at 18:00
  • OP, can you tell us why you are trying to sum the series and where you found the series? – Srivatsan Jul 24 '11 at 06:20
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    @Srivatsan: It's a far-fetched expression that shows up in a very particular problem that I was working on. The problem would not be very relevant for the series. In the end I was able to follow a completely different path and this series was no longer necessary. Since then I have done no further work on it. – a06e Jul 25 '11 at 19:32
  • @Patrick, your argument shows that the sum is $\le p^2/(1-p)$. By a similar one, it is also $\ge p^2/(1+p)$. – Did Sep 24 '11 at 16:45

2 Answers2

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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 $$ \sum_{n=1}^{\infty }\frac{1}{n}\left(p^{n+1}\right) = p \sum_{n=1}^{\infty } \frac{1}{n}\left(p^{n}\right) =p\int \sum_{n=1}^{\infty} \ p^{n-1} dp =p\int \frac{1}{1-p} dp = -p\ln(1-p) $$

saeed sani
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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.

Ross Pure
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