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.

Possible Duplicate:
Self-Contained Proof that $\sum\limits_{n=1}^{\infty} \frac1{n^p}$ Converges for $p > 1$

How would I go about showing that

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

exists for all $r>1$?

share|improve this question

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Do you know about the integral test? If so, you can use that one. –  Thomas Apr 11 '12 at 13:37
5  
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

2 Answers 2

up vote 0 down vote accepted

Here is how you can use the Integral Test.

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.

share|improve this answer

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).

share|improve this answer

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