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for $1, prove the p-series: $\sum_{n=1}^{\infty}n^{-p}$ is convergent.

please use Cauchy Rule (edit: that is, by showing directly that the sequence of partial sums is a Cauchy sequence) instead of concerned with integral and differential knowledge.

I have seen a proof before, without concern any tests about series. But I forgot it.

I had learned the condensation test. But the proof I had seen is not constructed by proved the condensation test, more precisely, not by shown every specify partial sums of a $p$-series is bounded in the corresponding items of a geometry series.

Edit. So much grateful to user6312 and Didier for your warmhearted doings.

For some reason however, I had to bother you two although I would wish have not to. User6312's second answer seem as a proof without importing differential or integral point but, the base idea, showing EACH $p$-series (here $p=2^{-k}$) was a telescope series SURELY could be done, came from the "middle point theorem". What do you think? For Didier's proof, like user6312's denote, a geometry series behind it, since you had to recurring to get the general term's formula of $A_n$.

I recalled the ever seen proof a little last night. That is like assemble from your two answers:

$\forall p\in(1,2)$, $\exists k\in\mathbb{N}$ so that $2^{-k}, then show the cutted-partial-sum (that is, summary from $N$ to $N+m$) of $2^{-k}$-series yields the SUM (multiply a constant factor) of $2^{1-k}$-series (if it's convergent), $\ldots\,$. At last (after finite steps), show the $2^{-1}$-series is a telescope series. Since the cutted-partial-sum of given $p^{-k}$-series yields the cutted-partial-sum of $2^{-k}$-series is obvious (to consider orresponding term), we have done. But the detail I haven't thought through over clearly. Might someone write it down?

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    What is the "Cauchy Rule"?2011-04-18
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    Cauchy Rule is sequence a_n convergent if and only if $|a_m-a_n|<\epsilon$ when $m>N(\epsilon)$, $n>N(\epsilon)$.2011-04-28
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    That is, you want a proof that the sequence of partial sums is a Cauchy sequence.2011-04-28
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    @Arturo Yes of course.2011-04-28
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    Relevant: http://math.stackexchange.com/questions/29450/self-contained-proof-that-sum-limits-n-1-infty-frac1np-converges-for/29466#294662016-09-25

3 Answers 3