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I have an exam in Calc 2 coming up. As such, I am doing previous exams given by our current professor. However, the exams lack a solution set, so I will post the question, and the answer I wrote down while practicing here. I am asking you to give the comments you would give a student taking an exam of yours, in addition to grade the answer on a scale from $0$ to $5$.

Show that $$\sum_{n=1}^{\infty}n^{-p}$$

converges for $p > 1$ and diverges for $p \le 1$.

My answer:

Let $p = 1$. We are now working with the well-known harmonic series: $$\sum_{n=1}^{\infty}\frac{1}{n}$$ Consider the integral $\int_1^{\infty}\frac{1}{x}\:dx$. We have that $$\int_1^{\infty}\frac{1}{x}\:dx = \lim_{L \to \infty}\int_1^L\frac{1}{x}\:dx = \lim_{L \to \infty}[\ln|x|]^L_1 = \infty$$ By the Integral Test, the harmonic series diverges. It follows that the $p$-series diverges for $p < 1$. Let $p > 1$. Then we can apply the Integral Test again, to obtain $$\int_1^{\infty}x^{-p}\:dx = \lim_{L \to \infty}[\frac{1}{-p+1}x^{-p+1}]_1^L = \frac{1}{p-1}$$ which is a finite value, so we are done.

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You have this backwards -- a $p$-series diverges if $p\leq 1$ and converges if $p>1$. In particular, the harmonic series $\sum 1/n$ diverges. – user128390 May 31 '14 at 15:03
Ah, its just a typo. – Andrew Thompson May 31 '14 at 15:07
up vote 2 down vote accepted

First you have a typo in the question and the series $\sum_n n^{-p}$ converges for $p>1$ and diverges for $p\le1$ and not the contrary. Now for your answer, OK it's correct but I'm not sure if in your country you accept applying a theorem without precising the hypothesis i.e. in this particular example we should say: Since the function $x\mapsto x^{-p}$ is non negative and monotone increasing on the interval $[1,+\infty)$ then by the integral test we have such and such thing and I know this can vary from one country to another and so the grad that you ask depends on it.

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Ah, thanks! We actually need to state that the hypothesis, so thank you for that. – Andrew Thompson May 31 '14 at 15:39
You're welcome. – user63181 May 31 '14 at 15:55
Nicely answered! – amWhy Jun 2 '14 at 17:25

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