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Possible Duplicate:
Contest problem about convergent series

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.2012-11-20
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    @glebovg Note that negative primes are excluded. (Btw, I am not one of the down voters).2012-11-20
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    @AD. Negative integers can not be prime.2012-11-20
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    @glebovg Hence if there were primes, that statement would be redundant.2012-11-20
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    @glebovg, I'm afraid you are wrong: $\,-7\,$ is as prime as $\,2\,,\,5\,\,\,or\,\,\,7\,$ within the integers ring.2012-11-20
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    As opposed to what one comment mentioned in the already deleted answer (which, btw, didn't deserve that much downvotes, imo., though it wasn't correct), the sequence $\,\{p_n\}\,$ doesn't have to be "eventually increasing", as the sequence $$\{p_n\}\,\,,\,p_n\begin{cases}n&\text{n is not a power of}\,\,\,2\\{}\\n-10&\text{when}\,\,n=2^k\end{cases}$$2012-11-20
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    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).2012-11-20
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    The point was that it had to have some increasing terms- You had the inequality in the wrong direction2012-11-20
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    @DonAntonio We can violate many rules in algebra, but I prefer [this](http://mathworld.wolfram.com/PrimeNumber.html) definition. Otherwise, if $-7$ is prime then it is divisible by $-7$, $7$, $-1$, and $1$.2012-11-20
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    Exactly @glebovg: an element in a (commutative?) unitary ring is a prime element if $\,p\mid ab\Longrightarrow p\mid a\vee p\mid b\,$ , and in the integers rings this boils down to be divisible only by its associate elements and *by the ring's units*, which are precisely $\,\pm 1\,,\,\pm7\,$ i n the example of $\,-7\,$ , say.2012-11-20
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    I think this is not true, because $\sum\nolimits_{n = 1}^\infty {\tfrac{1}{{{n^2}}}}$ converges but $\sum\nolimits_{n = 1}^\infty {\left( {\tfrac{{{n^4}}}{{\sum\nolimits_{k = 1}^n {{k^2}}}}}\right)} $ diverges.2012-11-20
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    @glebovg you forgot to square the bottom sum. It converges to $21\pi/2-144\ln(2)$2012-11-20
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    Strange because according to Maple it diverges.2012-11-20
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    @glebovg see edit2012-11-20
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    Yes. Thanks for pointing that out.2012-11-20
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    We may *assume* that the sequence $p_n$ is increasing. For each $L>0$ there can be at only finitely many $n$ with $p_n and by absolute convergence, we may rearrange. Such a permutation can only *increase* the target sum, if I'm not mistaken.2012-11-20
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    Using three periods instead of one of the commands `\ldots`, `\dotso`, ... messes up the formatting around the surrounding operators.2012-11-20
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    I think this is a duplicate but I can't find the original.2012-11-20
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    Like @joriki said. I vaguely remember that this was asked as question of the month in some math department where they do that each month, and that a solution was on their website.2012-11-20
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    @joriki: You're remembering correctly: [Contest problem about convergent series](http://math.stackexchange.com/questions/200514/)2012-11-20
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    The problem is really interesting, since using the inequality proven in http://math.stackexchange.com/questions/214634/prove-that-sum-k-1n-frac2k1a-1a-2-a-k4-sum-k-1n-frac1a-k/223836#223836 it is possible to dramatically improve the bound given in the "official" answer relative to http://math.stackexchange.com/questions/200514/ - so, even if this is a duplicate question, I think it deserves an answer explaining the improvement.2012-11-21

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