# Convergence of $\sum_{n=1}^{\infty}\frac{1}{n^\alpha}$

I'm trying to prove the convergence of $$\sum_{n=1}^{\infty}\frac{1}{n^\alpha}$$ with $\alpha > 1$.

For $\alpha \geq 2$ I can use the comparison test ($\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges) so I'm missing $2>\alpha>1$ and I'm pretty much out of ideas.

If you could offer some advice I would very much appreciate it.

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use the integral test – Ittay Weiss Nov 12 '12 at 9:15
How do you know $\sum 1/n^2$ converges? Use the same proof for $\sum 1/n^\alpha$ when $\alpha>1$. – wj32 Nov 12 '12 at 9:20
@wj32 I used $\sum_{k=1}^n \frac{1}{k(k+1)} <\sum_{k=1}^n \frac{1}{k^2} <1+ \sum_{k=1}^n \frac{1}{k(k-1)}$ but I don't see how I could use this proof for any real exponent $\alpha$ other than $2$. – hauptbenutzer Nov 12 '12 at 12:04
@hauptbenutzer How do you know your two bounds converge? – Jack M Apr 13 '14 at 20:06
@JackM There is a partial fraction decomposition of the terms and using this decomposition you can see that the bounds converge. $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$ – a... Apr 13 '14 at 20:12

Sorry to dig this up, but I would like to add that there is another way to do it without integral comparison test.

Write $\alpha=1+\beta$. Then $\beta>0$. Then group the summands in terms of powers of two and estimate by a geometric series:

$$\sum_{n=3}^\infty \frac{1}{n^\alpha}=\underbrace{\frac{1}{3^\alpha}+\frac{1}{4^\alpha}}_{\le \frac{1}{2^\alpha}+\frac{1}{2^\alpha}=2\frac{1}{2^\alpha}=\frac{1}{2^\beta}}+\underbrace{\frac{1}{5^\alpha}+\frac{1}{6^\alpha}+\frac{1}{7^\alpha}+\frac{1}{8^\alpha}}_{\le 4\cdot \frac{1}{4^{\alpha}}=\frac{1}{4^\beta}}+\cdots\le \sum_{k=1}^\infty \frac{1}{2^{k\beta}}=\frac{2^{-\beta}}{1-2^{-\beta}}<\infty$$

Edit: This is the Cauchy condensation test as pointed out below.

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@DanielFischer: This is basically the Cauchy condensation test. – Your Ad Here Sep 21 '14 at 19:02
Yes, but adding the words (possibly with a link to wikipedia) to your answer would be a good idea. – Daniel Fischer Sep 21 '14 at 19:03
:D ${}{}{}{}{}$ – Daniel Fischer Sep 21 '14 at 19:06
This is also what I ended up doing :D – hauptbenutzer Jul 16 '15 at 17:38

$$\frac{1}{n^{\alpha}} \le \int_{n-1}^n \frac{1}{x^{\alpha}}dx$$ for $\alpha \gt 1$ so
$$\sum_{n=1}^{\infty}\frac{1}{n^\alpha} = 1 + \sum_{n=2}^{\infty}\frac{1}{n^\alpha} \le 1+\int_{1}^\infty \frac{1}{x^{\alpha}}dx = 1+ \frac{1}{\alpha-1}.$$

Since each term is positive and the sum is bounded above, the series is convergent.

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$\int_1^\infty\frac1{x^\alpha}\mathrm{d}x=\frac1{\alpha-1}$ – robjohn Nov 12 '12 at 9:32
@robjohn: thank you - now edited – Henry Nov 12 '12 at 9:56

As the sequence $\displaystyle<n^\alpha>_{n=1}^{\infty}$ is decresing we can use the integral test to check its convergence.

$\displaystyle\int_{1}^{\infty}\frac{1}{x^\alpha}dx=1$ for all $\alpha>1$.

Hence the series is convergent.

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$\int_1^\infty\frac1{x^\alpha}\mathrm{d}x=\frac1{\alpha-1}$ – robjohn Nov 12 '12 at 9:32