$\displaystyle{\sum_{n=1}^\infty\frac{1}{n^p}}$ converges if $p>1$ and diverges if $0<p\le 1$.
Hello, I am unsure if my proof below is enough for the problem posed above. Any help is highly appreciated. Thank you!
Proof: Let $S_n (p)$ be the nth partial sum of the p-series $\displaystyle\sum_{n=1}^\infty \frac{1}{n^p}$ .
$\textbf{Case I}$ Let $p>0,$ $$1- \frac{1}{2^p} + \frac{2}{2^p} S_n (p) < S_{2n} (p) < 1 + \frac{2}{2^p} S_n(p).$$
$\textbf{Case II}$ Let $p<0$, $$1 + \frac{2}{2^p} S_n(p) < S_{2n}(p) < 1 - \frac{1}{2^p} + \frac{2}{2^p} S_n (p).$$
As $S_{2n}(p) = 1 + \frac{1}{2^p} + \frac{1}{3^p} + \dots + \frac{1}{(2n)^p}$
$= 1 + [\frac{1}{2^p} + \frac{1}{4^p} + \dots + \frac{1}{(2n)^p}] + [\frac{1}{3^p} + \frac{1}{5^p} + \dots + \frac{1}{(2n-1)^p}].$
For $p>0$, $$S_{2n}(p) > 1 + \frac{1}{2^p} S_n(p) + [\frac{1}{4^p} + \frac{1}{6^p} + \dots + \frac{1}{(2n)^p}].$$
Thus, $S_{2n} (p) > 1 + \frac{1}{2^p} S_n (p) - \frac{1}{2^p} + \frac{1}{2^p} S_n (p) = 1 - \frac{1}{2^p} + \frac{2}{2^p} S_n (p).$
Also, $$S_{2n} (p) < 1 + \frac{1}{2^p} S_n (p) + [\frac{1}{2^p} + \frac{1}{4^p} + \dots + \frac{1}{(2n)^p}] = \frac{2}{2^p} S_n(p) + 1.$$
The p-series is divergent when $p\leq 1$, and in this case, $$lim_{n\to\infty} \frac{S_{2n}(p)}{S_{n}(p)} = \frac{2}{2^p}.$$ (1*)
The p-series is convergent for $p>1$, and in this case, $\frac{2^{p} - 1} {2^{p} - 2} \leq lim_{n\to\infty} S_n(p) \leq \frac{2^p}{2^{p} - 2}.$ (2*)
When $p<0$, the p-series is divergent since the general term does not converge to $0.$ So let's consider the case when $0 < p \leq 1$. Assume the p-series is convergent, that is $lim_{n\to\infty} S_n(p) = S(p)$. Letting $n$ go to $\infty$ $$1 - \frac{1}{2^p} + \frac{2}{2^p} S(p) = \frac{2^{p} - 1}{2^p} + \frac{2}{2^p} S(p) \leq S(p),$$ and from this inequality we have $$0 < \frac{2^{p} - 1}{2^p} \leq \frac{2^{p} - 2}{2^p} S(p) \leq 0$$, which is a contradiction. Thus the p-series is divergent when $p\leq 1$. We obtain $$lim_{n\to\infty} \frac{S_{2n}(p)} {S_n(p)} = \frac{2}{2^p}$$ by dividing both inequalities by $S_{n}(p)$ and letting $n$ go to $\infty$. This proves (1*).
Now, let $p>1$. From the inequality at start (Case I), we have $$S_{n}(p) < S_{2n} (p) <1 + \frac{2}{2^p} S_{n} (p),$$ and then $$0 < (1-(\frac{2}{2^p})S_{n}(p) <1.$$ Hence, $$S_{n} (p) < \frac{2^p}{(2^{p} - 2}$$ for all n, so the sequence $\{S_{n}(p)\}$ is bounded and increasing. Thus the limit, $lim_{n\to\infty} S_{n}(p)$ exists, and hence the p-series is convergent for $p>1$. The inequality is obtained by letting $n$ go to $\infty$ in the first part (Case I).
The following we have done is tantamount to $a_k=\displaystyle{\frac{1}{k^p}}$ so that $a_{2^k} = \displaystyle{\frac{1}{2^{kp}}}$ and hence that $2^k\,a_{2^k} = \displaystyle{\frac{2^k}{2^{kp}} = \frac{1}{2^{k(p-1)}} = (2^{(1-p)})^k}$. Therefore, $$\sum_{k=1}^\infty 2^k\,a_{2^k} = \sum_{k=1}^\infty (2^{(1-p)})^k$$ is a geometric series that converges if $2^{(1-p)}<1$ and diverges if $2^{(1-p)}>1$. Converges if $p>1$ and diverges if $0<p<1$.
This looks much more complicated than it needs to be. First off, the case $p \leq 0$ is taken care of by the limit test.
For $p>0$, the summands are decreasing, and you can compare to an appropriate integral of $x^{-p}$, and you know how to integrate $x^{-p}$. Which integral you want to compare to depends on whether you want to show convergence or divergence. If you want convergence, then you want an integral $I$ with $\sum_{n=1}^\infty n^{-p} \leq I$, which means that you want to treat this sum as a right hand rectangle approximation of $I$. If you want divergence then it goes the other way, and you want the sum to be a left hand rectangle approximation. All this left/right distinction really means is whether you are comparing to $\int_0^\infty x^{-p} dx$ or $\int_1^\infty x^{-p} dx$.
Probably this exact case is covered in whatever reference you are referring to, by the way, as it is a rather standard thing.