Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
This link [math.stackexchange.com/questions/29450/… could interest you. –  Davide Giraudo Jan 21 '12 at 22:03

2 Answers 2

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

share|improve this answer
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 $\S 4.3$ of these notes for a statement of CCT and its application to convergence of $p$-series.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.