How to show that $$\sum_{n=1}^{\infty} \frac{1}{n^k}$$ converges for all integer $k > 1$?
I know that comparison test would suffice to show that, but don't know how to start.
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2$\begingroup$ Note that it suffices to establish convergence for $k=2$. Then you can use the comparison test to conclude the same for larger $k$. So that reduces the number of cases you have to consider from countably infinite to one. :-) $\endgroup$– user169852Commented Sep 11, 2014 at 4:54
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$\begingroup$ I guess that 12 minutes are too short a time span to have the opportunity to note that the title should be heavily edited, if one is to follow the usual conventions of mathematical language. $\endgroup$– DidCommented Sep 11, 2014 at 5:34
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3 Answers
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Compare $\frac{1}{n^2}$ with $\frac{1}{n(n-1)}$ ($n \ge 2$).
The second of these decomposes into partial fractions and the infinite sum can be easily computed.
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Use this: Suppose $a_n\geq a_{n+1}\geq 0 \; \forall\; n\;\in\mathbb{N} $. Then the series $\sum_{n=1}^{\infty}a_n$ converges if and only if $\sum_{m=0}^{\infty}2^ma_{2^m}$ converges.
Here we get $\sum_{m=0}^{\infty}2^m\frac{1}{2^{mk}}=\sum_{m=0}^{\infty}2^{(1-k)m}$. Note that this ia a geometric series with commom ratio $2^{1-k}$. Since $k>1$, $2^{1-k}<1$. Thus the series $\sum_{m=0}^{\infty}2^m\frac{1}{2^mk}$ converges. Hence the given series convergent.
answered Sep 11, 2014 at 5:31
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1$\begingroup$ Note: this very much requires the sequence to be monotone! $\endgroup$ Commented Sep 11, 2014 at 6:06
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$\begingroup$ ya, monotone is needed, i edited. Thank you. $\endgroup$ Commented Sep 11, 2014 at 7:02