How to prove convergence of $\sum_{n=1}^{\infty}n^{-k}$?

Question

I want to check convergence or divergence of $\displaystyle \sum_{n=1}^{\infty}\dfrac{1}{n^k}$

I already proved that it converges for $k\geq 2$ and diverges for $0\leq k \leq 1$, but i don't know how to prove convergence for $1 < k < 2$, any ideas?

Edit: no integrals allowed

Answer 1

Applying Cauchy's condensation test, you have that, for $\alpha>0$, $\sum\limits_{k=1}^\infty k^{-\alpha}$ is convergent if and only if $\sum_{k=0}^\infty 2^k 2^{-k\alpha}$ is convergent.

But $$\sum_{k=0}^n 2^{(1-\alpha)k}=\frac{1-2^{(n+1)(1-\alpha)}}{1-2^{1-\alpha}}$$ which coverges as $n\to\infty$ if and only if $1-\alpha<0$.

Answer 2

Bernoulli's Inequality says that, for $k\gt1$, $$ \begin{align} (n+1)^k\left(\frac1{n^{k-1}}-\frac1{(n+1)^{k-1}}\right) &=n\left(1+\frac1n\right)^k-(n+1)\\ &\ge n\left(1+\frac kn\right)-(n+1)\\[6pt] &=k-1 \end{align} $$ Therefore, $$ \frac1{(n+1)^k}\le\frac1{k-1}\left(\frac1{n^{k-1}}-\frac1{(n+1)^{k-1}}\right) $$ Thus, $$ \begin{align} \sum_{n=1}^\infty\frac1{n^k} &=1+\sum_{n=1}^\infty\frac1{(n+1)^k}\\ &\le1+\sum_{n=1}^\infty\frac1{k-1}\left(\frac1{n^{k-1}}-\frac1{(n+1)^{k-1}}\right)\\ &=\frac k{k-1} \end{align} $$