# Showing a series is convergent. [duplicate]

Possible Duplicate:

Let ${p}_{n}\in \mathbb{R}$ be positive for every $n$ and $\sum_{n=1}^{∞}\cfrac{1}{{p}_{n}}$ converges,

How do I show that $\sum_{n=1}^{∞}{p}_{n}\cfrac{{n}^{2}}{{({p}_{1}+{p}_{2}+\dotso+{p}_{n})}^{2}}$ converges?

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I hope $p_n$ is not the $n$-th prime, otherwise your premise is false because the sum of the reciprocal of the primes is divergent. –  glebovg Nov 20 '12 at 17:16
@glebovg Note that negative primes are excluded. (Btw, I am not one of the down voters). –  AD. Nov 20 '12 at 17:27
He says "Let $p_n$ be positive for every $n$", I think $p_n$ is just a sequence that fullfils this (and $\sum^\infty_{n=1}\frac 1 {p_n}$ converges). –  Stefan Nov 20 '12 at 17:54
I think this is a duplicate but I can't find the original. –  joriki Nov 20 '12 at 21:49
@joriki: You're remembering correctly: Contest problem about convergent series –  user50154 Nov 20 '12 at 23:51

## marked as duplicate by Lukas Geyer, Leitingok, Martin Sleziak, Did, ShikariShambuNov 21 '12 at 16:57

Assume this fact $(\clubsuit)$: Prove that $\sum_{k=1}^n \frac{2k+1}{a_1+a_2+...+a_k}<4\sum_{k=1}^n\frac1{a_k}.$.
If you define $P_N=\sum_{n=1}^N p_n\,,\; C=\sum_{n=1}^{+\infty}\frac{1}{p_n}$ and $S_N=\sum_{n=1}^{N}\frac{n^2 p_n}{P_n^2}$ you have:
$$S_N < \frac{1}{p_1} + \sum_{n=2}^{N}\frac{n^2(P_n-P_{n-1})}{P_n P_{n-1}} = \frac{5}{p_1}+\sum_{n=2}^{N-1}\frac{2n+1}{P_n}-\frac{N^2}{P_N},$$
so, in virtue of $(\clubsuit)$, you have:
$$S_N < \frac{2}{p_1} + 4C.$$