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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? –  M. Strochyk Sep 27 '12 at 7:55
yes but i want to find p such that the sum converges to 1 –  Wuschelbeutel Kartoffelhuhn Sep 27 '12 at 7:56
@M.Strochyk if $p>1$ this is the $\zeta$ function... –  draks ... Sep 27 '12 at 7:57
@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 $N>1$ –  M. Strochyk Sep 27 '12 at 8:00
LOL, you're of course right. (How could I miss this...) –  Wuschelbeutel Kartoffelhuhn Sep 27 '12 at 8:08
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2 Answers

up vote 0 down vote accepted

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... –  draks ... Sep 27 '12 at 8:12
That is true, I wil edit that in. –  An.Ditlev Sep 27 '12 at 8:15
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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 x –  student101 Sep 27 '12 at 8:01
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