Does the series $$\sum_{n=1}^\infty \frac{1}{n^p}$$ converge for $p\in [1, +\infty)$? If it does not, how to prove it?
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Do you know the [Integral Test](http://tutorial.math.lamar.edu/Classes/CalcII/IntegralTest.aspx)? – David Mitra Feb 12 '13 at 17:53
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$p>1$ is handled in http://math.stackexchange.com/questions/29450/self-contained-proof-that-sum-limits-n-1-infty-frac1np-converges-for. For $p=1$, search this site or the internet generally for "harmonic series". – Jonas Meyer Feb 12 '13 at 17:54
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For $p = 1$, this is the harmonic series which is divergent:
$$1 + \frac12 + \underbrace{\frac13 + \frac14}_{>2\cdot \frac12} + \underbrace{\frac15 + \cdots + \frac 18}_{> 4\cdot \frac18 = \frac12} + \underbrace{\frac19 + \cdots + \frac{1}{16}}_{> 8 \frac{1}{16} = \frac12} + \cdots $$
and from this it's clear that the partial sums of the series are unbounded, i.e. the series diverges.
For $p > 1$, convergence follows most easily from the integral test.
mrf
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No, the series converges only for $p\in (1,\infty)$, as can be shown quite easily using the integral test. The case $p=1$ yields the harmonic series which is well-known to diverge as can be shows in various ways.
Ittay Weiss
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This is a standard Analysis question. It's answered in say Baby Rudin (Principles of Mathematical analysis. You can use the following from Rudin:
If $x_n$ is decreasing and non negative, $\sum x_n$ converges IFF $\sum 2^nx_{2^n}$ converges.
Gautam Shenoy
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