$\sum_{n=1}^\infty \frac{1}{n^p}$ converges for any $p>1$ I read that the harmonic series $\displaystyle\sum_{n=1}^\infty\dfrac{1}{n}$ diverges but for any $p>1$ the series $\displaystyle\sum_{n=1}^\infty\dfrac{1}{n^p}$ converges. I found this quite difficult to comprehend and could not find a proof by myself. Is there a nice proof of this?
 A: Consider the series $\sum a_n $ where $(a_n)$ is decreasing and positive. Also, consider the series $\sum 2^n a_{2^n}$. Consider their partial sums $(s_n), (t_n)$
Note if $n < 2^n$, then $s_n = a_1 + a_2 + ... + a_n \leq a_1 + (a_2+a_3) + (a_4 + a_5 + a_6 + a_7) + ... + (a_{2^n} + ... + a_{2^{n+1}-1}) \leq a_1 + 2a_2 + ... + 2^n a_{2n} = t_n \implies \boxed{ s_n \leq t_n }$
If $2^n < n$, then $s_n  \geq a_1 + a_2 + (a_3 + a_4) + ... + (a_{2^{n-1}+1} + ... + a_{2^n} ) \geq \frac{1}{2} a_1 + a_2 + 2 a_4 + ... + 2^{n-1}a_n = \frac{1}{2} t_n \implies \boxed{ 2s_n \geq t_n }$
Since series of nonnegative terms converges iff its sequence of partial sums are bounded, then by the work above we obtain the following
$$ \boxed{ \sum a_n \; \; \; \mathbf{Converges \; \; iff} \; \; \; \sum 2^n a_{2^n} \; \; \; \mathbf{Converges} }$$
Thus, with $a_n = \frac{1}{n^p}$, then $a_{2^n} = \frac{1}{2^{np}}$, thus
$$ \sum 2^n \frac{1}{2^{np}} = \sum \left( \frac{1}{2} \right)^{n(p-1)} = \sum \left( \frac{1}{2^{p-1}} \right)^n$$
Clearly this series $\mathbf{converges}$, being a geometric series, iff $p-1>0 $ iff $p > 1$ cause otherwise it would a divergent geometric series.
A: One can use the integral test to ascertain the convergence of $\displaystyle\sum_{n=1}^\infty\dfrac{1}{n^p}$ where $p>1$. (link to integral test: https://en.wikipedia.org/wiki/Integral_test_for_convergence)
Let $p>1$ be a real number. Let $f:[1,\infty)\to \mathbb{R}$ be a function defined as $f(x):=1/x^p$. The function $f$ is non-negative, monotone decreasing and continuous. By the integral test for convergence, the series$$\displaystyle\sum_{n=1}^\infty f(n)$$ converges if and only if the integral $$\displaystyle\int_1^\infty f(x)\mathrm{dx}$$ converges.
We have $$\displaystyle\int_1^\infty f(x)\mathrm{dx}=\displaystyle\int_1^\infty\dfrac{1}{x^p}\mathrm{dx}=\displaystyle\lim_{a\to\infty}\int_1^a\dfrac{1}{x^p}\mathrm{dx}$$
But we have $$\displaystyle\lim_{a\to\infty}\int_1^a\dfrac{1}{x^p}\mathrm{dx}=\displaystyle\lim_{a\to\infty}\left.\dfrac{x^{1-p}}{1-p}\right|_1^a=\displaystyle\lim_{a\to\infty}\left(\dfrac{a^{1-p}}{1-p}-\dfrac{1}{1-p}\right)=\displaystyle\lim_{a\to\infty}\dfrac{a^{1-p}}{1-p}-\displaystyle\lim_{a\to\infty}\dfrac{1}{1-p}$$
Since $1-p<0$, we have $$\displaystyle\lim_{a\to\infty}\dfrac{a^{1-p}}{1-p}=0$$
Thus $$\displaystyle\int_1^\infty f(x)\mathrm{dx}=\dfrac{1}{p-1}$$, and the integral $\displaystyle\int_1^\infty f(x)\mathrm{dx}$ converges.
By the integral test the series $\displaystyle\sum_{n=1}^\infty\dfrac{1}{n^p}$ also converges.
