# How to show the sum of $\frac1{n^r}$ exists for all $r > 1$? [duplicate]

How would I go about showing that

$$\sum_{n=1}^{\infty } \frac{1}{n^{r}}$$

exists for all $r>1$?

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## marked as duplicate by Ragib Zaman, Asaf Karagila, Paul, t.b., Zev ChonolesApr 12 '12 at 2:41

This question was marked as an exact duplicate of an existing question.

Do you know about the integral test? If so, you can use that one. – Thomas Apr 11 '12 at 13:37
Here's a nice proof. – Ragib Zaman Apr 11 '12 at 13:38
I'm not sure this should be closed as a duplicate (at least not for the proposed "duplicate"), as this question would allow the integral test to be used, and illustrated, in the answer. – David Mitra Apr 11 '12 at 13:42

Let $f(x) = \frac{1}{x^r}$ and $r > 1$. Note that $f$ is continuous, positive, and decreasing on the interval $[1, \infty)$. Note also that for an integer $n$, you have $f(n) = \frac{1}{n^r}$. So the Integral Test applies. Hence you need to prove that the integral $$\int_1^{\infty} f(x) dx$$ is convergent. You have \begin{align}\int_1^{\infty}f(x) dx &= \lim_{N\to \infty}\int_1^{N} x^{-r} dx\\ &= \lim_{N \to \infty} \left[\frac{1}{1-r}x^{1-r}\right]_1^N \\ &= \lim_{N\to \infty} \frac{1}{1-r}\left[\frac{1}{N^{r-1}} - 1\right] \\ &= \frac{1}{r-1}. \end{align} We used here that $r > 1$ so $r-1 > 0$, and so $$\lim_{N\to \infty} \frac{1}{N^{r-1}} = 0.$$ So the integral is convergent, so by the Integral Test, the series is convergent.
For $r>1$ the function $\frac{1}{x^r}$ is positive and decreases in the interval $[1, \infty)$ then the series converges since the improper integral from 1 to infinite converges. (You can verify the integral convergence because you can find a primitive of the function).