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I wanted to know if there is an online reference I can use to find out known results about convergent series. I could not find this one, for example, on wikipedia

$\sum_{k=1}^{+\infty} \left(\frac{1}{2}\right)^k k^2$

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No, but there is a list of tests for convergence, which would be much more useful for someone taking a test on this topic... –  Antonio Vargas May 12 '12 at 23:01
    
Are you trying to work out whether it converges, or are you trying to work out what it converges to? –  Gerry Myerson May 13 '12 at 12:17

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up vote 4 down vote accepted

There exists an $N$ such that for all $k > N$, $k^2 \le (3/2)^k$. This is just because $$ \lim_{k \to \infty} \frac{(3/2)^k}{k^2} = \lim_{k \to \infty} \left( \frac{\sqrt{3/2}^k}{k} \right)^2 = \infty, \quad \Longrightarrow \quad \exists N \mbox{ s.t.} \frac{(3/2)^k}{k^2} \ge 1 \mbox{ for all $k<N$}. $$ Therefore, $$ \sum_{k=N+1}^{\infty} \frac{k^2}{2^k} \le \sum_{k=N+1}^{\infty} \frac{(3/2)^k}{2^k} = \sum_{k=N+1}^{\infty} \left( \frac 34 \right)^k $$ which converges.

If you're looking for some library which lists all known patterns of convergence series, you're not trying to understand the mathematics behind those series correctly. The right way to study this (for example, for an exam) is to not only understand the tricks used to show that one series converges/diverges, but also to understand why this was the chosen trick and that it wasn't done purely by luck. Also I suggest trying to apply every elementary tests a few times to know when they work and when they don't.

Hope that helps,

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Without saying more about $N$, your sums should start with $k=N+1$, no? –  Matthew Conroy May 13 '12 at 3:47
    
@Matthew : Indeed, I'll correct that.... heh. –  Patrick Da Silva May 13 '12 at 12:11
    
btw, the result is 2 –  ACAC May 14 '12 at 4:42
    
@Bob : Sure, you could also compute this series very easily. I didn't bother. –  Patrick Da Silva May 15 '12 at 1:53

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