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$\begingroup$

$\frac{1}{x^n}$

Consider an infinite series like this where x if defined for the natural numbers and n is fixed.

I know that when n = 1 the series diverges (harmonic series), and for n=2 I found a website that said it converges into $\pi^2/6$. Is there an easy way to find the value of n required to make the series converge into 1?

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    Did you mean series of kind $\sum\limits_{n=1}^{\infty}{\frac{1}{n^P}},\quad $, where $p$ is fixed, so called [*P-series*](http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#P-series)?2012-09-27
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    yes but i want to find p such that the sum converges to 12012-09-27
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    @M.Strochyk if $p>1$ this is the $\zeta$ function...2012-09-27
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    @Wuschelbeutel Kartoffelhuhn This is series with positive terms, and the first term is equal to $1$, so any partial sum $\sum\limits_{n=1}^{N}{\frac{1}{n^p}}>1$ for $N>1$2012-09-27
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    LOL, you're of course right. (How could I miss this...)2012-09-27

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There is a result which states that $\sum_{n=1}^\infty1/n^{2k}=\frac{(-1)^k-1 ( 2^{2k} B_{2k}\pi^{2k} )}{2(2k)!}$. Where $B_k$ is the $k'th$ bernouilli number.

In general you have $\sum_{n=1}^\infty1/n^{s}=\zeta(s)$ for $Real(s)>1$, where $\zeta(s)$ is the Riemann zeta function.

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    I think, the $\zeta$ function can only represented by the sum, if the real part of $s$ is larger than 1...2012-09-27
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    That is true, I wil edit that in.2012-09-27
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if n>1 then it converges; the series diverges for other values of n

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    how about trying the geometric series sum? for appropriate values of x2012-09-27