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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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Leitingok
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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 - Stop Putin - Nov 20 '12 at 17:27
  • @AD. Negative integers can not be prime. – glebovg Nov 20 '12 at 17:28
  • @glebovg Hence if there were primes, that statement would be redundant. – AD - Stop Putin - Nov 20 '12 at 17:31
  • @glebovg, I'm afraid you are wrong: $,-7,$ is as prime as $,2,,,5,,,or,,,7,$ within the integers ring. – DonAntonio Nov 20 '12 at 17:47
  • 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 $${p_n},,,,p_n\begin{cases}n&\text{n is not a power of},,,2\{}\n-10&\text{when},,n=2^k\end{cases}$$ – DonAntonio Nov 20 '12 at 17:52
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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). – Stefan Nov 20 '12 at 17:54
  • The point was that it had to have some increasing terms- You had the inequality in the wrong direction – blitzer Nov 20 '12 at 17:56
  • @DonAntonio We can violate many rules in algebra, but I prefer this definition. Otherwise, if $-7$ is prime then it is divisible by $-7$, $7$, $-1$, and $1$. – glebovg Nov 20 '12 at 17:58
  • 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. – DonAntonio Nov 20 '12 at 18:15
  • 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. – glebovg Nov 20 '12 at 18:16
  • @glebovg you forgot to square the bottom sum. It converges to $21\pi/2-144\ln(2)$ – Jean-Sébastien Nov 20 '12 at 18:21
  • Strange because according to Maple it diverges. – glebovg Nov 20 '12 at 18:25
  • @glebovg see edit – Jean-Sébastien Nov 20 '12 at 18:26
  • Yes. Thanks for pointing that out. – glebovg Nov 20 '12 at 18:31
  • We may assume that the sequence $p_n$ is increasing. For each $L>0$ there can be at only finitely many $n$ with $p_n<L$ and by absolute convergence, we may rearrange. Such a permutation can only increase the target sum, if I'm not mistaken. – Hagen von Eitzen Nov 20 '12 at 18:48
  • Using three periods instead of one of the commands \ldots, \dotso, ... messes up the formatting around the surrounding operators. – joriki Nov 20 '12 at 21:44
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    I think this is a duplicate but I can't find the original. – joriki Nov 20 '12 at 21:49
  • 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. – Did Nov 20 '12 at 22:29
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    @joriki: You're remembering correctly: Contest problem about convergent series –  Nov 20 '12 at 23:51
  • 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. – Jack D'Aurizio Nov 21 '12 at 16:16

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