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.
Asked
Active
Viewed 123 times
0
-
2Note 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. :-) – Sep 11 '14 at 04:54
-
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. – Did Sep 11 '14 at 05:34
3 Answers
5
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.
user164587
- 1,529
0
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.
