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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

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    My guess is that there is no nice closed form solution.2011-06-02
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    Certainly the $n^{n+1}$ factor will likely preclude an expression in terms of currently known functions...2011-06-02
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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.2011-06-02
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    If you define this as function $f(p)$ then the terms of the powerseries begin with $ p^2 - 3/2*p^3 + 31/12*p^4 - 31/9*p^5 + 10007/2160*p^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 $|x|< 1$ )2011-06-02
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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...)2011-06-02
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    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.2011-06-03
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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').2011-07-10
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    You should somehow use that $$\Bigl({1\over 1+{p\over n}}\Bigr)^n\doteq e^{-p}\ .$$2011-07-10
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    OP, can you tell us _why_ you are trying to sum the series and _where_ you found the series?2011-07-24
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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.2011-07-25
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    @Patrick, your argument shows that the sum is $\le p^2/(1-p)$. By a similar one, it is also $\ge p^2/(1+p)$.2011-09-24

2 Answers 2

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If $0

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    Your point being?2011-09-24
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    I don't know why the site can not loaded completely.If you go to edit page ,you can see the complete my answer. Thanks2011-09-24
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    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...2011-09-24
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    it's good. my answer result a Upper bound for summation that it is very near real answer.2011-09-24
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    @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).2011-09-24
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    thanks didier, but I think my answer was correct and you did brief it.2011-09-24
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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.

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    Your first equation is not correct, because $$(n^{-1})^{n+1}\neq n^n.$$ Instead, $$(n^{-1})^{n+1}=n^{-(n+1)}=n^{-n-1}.$$2011-07-10
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    Also, the geometric series formula only works if $r$ is fixed, in your case $r = \frac {n^{-1}p}{p+n}$ depends on $n$.2011-07-10
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    I thought there was something wrong, probably shouldn't do maths that late. Sorry guys.2011-07-11
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    Also, I can take down the answer if you want.2011-07-11
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    Don't do that .2011-10-01