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Let r > 1 be a real number. Prove that the following series is convergent. $$\sum_{n = 1}^{\infty}\frac{1}{n^r}$$

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marked as duplicate by Martin, Git Gud, Did, Amzoti, Start wearing purple May 23 '13 at 21:05

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.

I tried comparison, ratio and root tests but none of them help. – Otávio Rapôso May 23 '13 at 20:34
Hint: Open any book about series. – Git Gud May 23 '13 at 20:34
I don't have any here now. I'm searching in internet, but I guess here is the fastest way. – Otávio Rapôso May 23 '13 at 20:40
You want to sum from $n =1$, not $n=0$. – Martin May 23 '13 at 20:42
It might be helpful to note that $1/n^r = 1 \cdot 1/n^r$. Which is a rectangle of sides of length $1$ and $1/n^r$ respectively. If you draw these rectangles on the plane you will get an idea of how to bound it from above. (Also draw $f(x)=1/x^r$ on the same picture) – Rudy the Reindeer May 23 '13 at 20:50

You can use the integral test.

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Back to basics. The partial sums are clearly increasing. So we only need to show they are bounded above. Consider the partial sum up to $n=2^{k}-1$.

Look at the first term, the sum of the next $2$ terms, the sum of the $4$ terms after that, the sum of the $8$ terms after that, and so on.

The first term is $1$.

The sum of the next $2$ terms is $\lt \frac{2}{2^r}=\frac{1}{2^{r-1}}$.

The sum of the next $4$ terms is $\lt \frac{4}{2^{2r}}=\frac{1}{2^{2(r-1)}}$.

The sum of the next $8$ terms, by a similar argument, is $\lt \frac{1}{2^{3(r-1}}$.

And so on. So the sum of the terms up to $2^k-1$ is less than $$1+\frac{1}{2^{r-1}}+\frac{1}{2^{2(r-1)}}+\cdots+\frac{1}{2^{(k-1)(r-1)}}.$$ This is a finite geometric series, with sum equal to $$\frac{1-2^{-k(r-1)}}{1-2^{-(r-1)}}.$$ In particular, it has the upper bound $\dfrac{1}{1-2^{-(r-1)}}$.

Remark: Without mentioning it explicitly, we have used the idea behind the Cauchy Condensation Test.

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