p-series

P- series

$$\sum k^{-p}=\sum _{k=1}^{\infty }\frac{1}{k^p}\{\begin{array}{ll}\text{converges}& \text{if }p>1\\ \text{diverges}& \text{if }p\le 1\end{array}$$

The harmonic series is the boundary case $p=1$.
Smaller $p$ diverges faster (terms shrink slower), larger $p$ converges (terms shrink fast enough).

Proof:

Using the cauchy condensation test, we see the sum is convergent iff the following geometric series converges:

$$\sum _{k=0}^{\infty }{2}^k\frac{1}{(2^k{)}^p}=\sum _{k=0}^{\infty }{2}^{(1-p)k}=\sum _{k=0}^{\infty }(2^{1-p}{)}^k=\sum _{k=0}^{\infty }{q}^k\text{ with }q=2^{1-p}$$

which converges iff $|q|<1⟺2^{1-p}<1⟺p>1$. $\square$

For what values of $c$ does $\sum \frac{1}{k^c}$ converge?

Converges if $c>1$, diverges if $c\le 1$. The harmonic series is the boundary case $c=1$.

Does ${\sum }_{k=1}^{\infty }\frac{1}{k^2}$ converge?

Disregarding the fact that we know it’s a p-series with $p>1$ and thus converges, we can approach it like this:

$$\begin{array}{rl}\sum _{k=1}^{\infty }\frac{1}{k^2}=\sum _{k=1}^{\infty }\frac{1}{k^2}\cdot \frac{k+1}{k+1}=\sum _{k=1}^{\infty }\frac{1}{k(k+1)}\cdot \frac{k+1}{k}& \le 2\sum _{k=1}^{\infty }\frac{1}{k(k+1)}\\ & =2\sum _{k=1}^{\infty }\left(\frac{1}{k}-\frac{1}{k+1}\right)\\ & =2\left(1-0\right)\\ & =2<\infty \end{array}$$

Applied partial fractions to get a telescoping sum. The series is bounded above by $2$, so it converges.