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Show that if $p>1$, $\sum\frac{1}{n^{p}}$ converges and if $p<1$ it diverges for $p\in\mathbb{R}^{+}$.

Is there any way to show another series converges or diverges and then use the Comparison Test to prove this?

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This link [… could interest you. – Davide Giraudo Jan 21 '12 at 22:03

4 Answers 4

You can show this using the integral test: $\sum \frac{1}{n^p} $ behaves as $\int_1^\infty dx/x^p$ which diverges if $p \leq 1$.

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Why does it diverge if $p\leq1$? – Emir Jan 21 '12 at 22:48
$\int_1^N \frac{dx}{x}=\log(N)$. This blows up (slowly) as $N\to\infty$. Then you can do $p<1$ in a similar way, or by comparison with the case $p=1$. – André Nicolas Jan 22 '12 at 1:04

There are many ways to skin this particular cat. Among methods not already mentioned, perhaps my favorite is the use of Cauchy's Condensation Test. See $\S 2.4.3$ of these notes for a statement of CCT and its application to convergence of $p$-series.

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I don't see a section 4.3 in the notes. – Suzu Hirose Dec 31 '14 at 1:21
@SuzuHirose ?? Starting on page 48. – Did Dec 31 '14 at 18:07

Lemma 1 (Cauchy criterion). Let $(a_n)_{n=1}^\infty$ be a decreasing sequence of non-negative real numbers (so $a_n\ge0$ and $a_{n+1}\le a_n$ for all $n\ge 1$). Then the series $\sum_{n=1}^\infty a_n$ is convergent if and only is the series $$\sum_{k=0}^\infty2^ka_{2^k}=a_1+2a_2+4a_4+8a_8+\dotsb$$ is convergent.

Applying this lemma we can prove:

Proof. The sequence $(1/n^p)_{n=1}^\infty$ is non-negative and decreasing (why?), and so the Cauchy criterion applies. Thus this series is convergent if and only if $$\sum_{k=0}^\infty2^k\frac{1}{(2^k)^p}$$ is convergent. But by the laws of exponentiation we can write this as the geometric series (why?)$$\sum_{k=0}^\infty(2^{1-p})^k.$$

The geometric series $\sum_{k=0}^\infty x^k$ converges if and only if $|x|<1$ (why?). Thus the series $\sum_{n=1}^\infty1/n^p$ will converge if and only if $|2^{1-p}|<1$, which happens if and only if $p>1$ (why? Try to proving it just using the laws of exponentiation, and without using logarithms).

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This is a great question.

Consider $S$

$$S = \sum_{n=1}^{\infty} \frac{1}{n^p}$$

Where $p$ is even, $p > 1$

Consider $f(z) = 1/z^p$ such that $z^p = 0 \implies z = 0$

So $z$ is a pole. By the residue theorem,

$$\sum_{n=-\infty}^{\infty} \frac{1}{n^p} = -\sum \text{sum of residues of} \space \pi\cot(\pi z) f(z)$$

Because $p$ is even,

$$\sum_{n=-\infty}^{\infty} \frac{1}{n^p} = \sum_{n=1}^{\infty} \frac{1}{n^p} + \sum_{n=-\infty}^{1} \frac{1}{n^p} = 2\sum_{n=1}^{\infty} \frac{1}{n^p}$$

The residues are:

$$\lim_{z \to 0} \frac{\pi\cot(\pi z)}{z^{p-1}} = 2\sum_{n=1}^{\infty} \frac{1}{n^p} $$

Which exists for $p > 1$, even and you see that the denominator grows too slowly if $p < 1$ which means that limit wont exist.

This however only works in my knowledge for $p$ even.

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