$u_n=\sum_{i=1}^{n} \frac{1}{i^\beta}$ converge

Question

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Self-Contained Proof that $\sum\limits_{n=1}^{\infty} \frac1{n^p}$ Converges for $p &gt; 1$

We know that $\lim u_n=\sum_{i=1}^{n} \frac{1}{i} =\infty $. But I can't prove that with $\beta >1$: $u_n=\sum_{i=1}^{n} \frac{1}{i^\beta}$ converge

Do you know the integral test? – Brian M. Scott Jan 07 '13 at 04:31 — Brian M. Scott, Jan 07 '13 at 04:31
Here is a beautiful argument by joriki. – Jan 07 '13 at 04:31 — , Jan 07 '13 at 04:31

score 0 · Answer 1 · answered Jan 07 '13 at 04:36

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Hint: Let $U = u_{\infty}$. Think about the relationship between $iU$ and $U$.

answered Jan 07 '13 at 04:36

Arthur Small

1 Answers1