# Thread: So, how exactly does a p-series converge

1. ## So, how exactly does a p-series converge

I'm studying the p-series convergence test right now , I understand what the theorem says but I don't understand why it works. I understand that a harmonic series [sum(1/k)] diverges and I understand why it diverges. But to me the p-series [sum(a/k^p)] is basically the same kind of series but the terms will just approach zero a bit faster.

2. Say the p-series is

$\displaystyle \sum_{k=1}^{\infty}\frac{a}{k^p}.$

Imagine the terms of the p-series drawn as rectangles on the positive $\displaystyle x$-axis. These rectangles can be shifted to fall under the curve defined by

$\displaystyle y=\frac{a}{x^p}.$

Therefore, the total sum of the area of these rectangles -- that is, the value of the sum $\displaystyle \sum_{k=2}^{\infty}\frac{a}{k^p}$ -- is less than the value of the integral

$\displaystyle \int_1^{\infty}\frac{a}{x^p}\,dx.$

Because this integral is finite (for $\displaystyle p>1$), the series converges.

Edit: Replaced $\displaystyle k=1$ with $\displaystyle k=2$ in the above expression.

3. Originally Posted by dmoreau
I'm studying the p-series convergence test right now , I understand what the theorem says but I don't understand why it works. I understand that a harmonic series [sum(1/k)] diverges and I understand why it diverges. But to me the p-series [sum(a/k^p)] is basically the same kind of series but the terms will just approach zero a bit faster.

Besides the integral test you can use Cauchy's Condensation Test: if a sequence $\displaystyle \{a_n\}_{n=1}^\infty$ is positive and monotone converging to zero, then the series $\displaystyle \sum\limits_{n=1}^\infty a_n\,\,and\,\,\sum\limits_{n=1}^\infty 2^na_{2^n}$ converge and diverge together.

So take the p-series $\displaystyle \sum\limits_{n=1}^\infty\frac{a}{n^p}$ . By CCT, this series converges iff converges the series

$\displaystyle \sum\limits_{n=1}^\infty\frac{2^na}{2^{np}}=a\sum\ limits_{n=1}^\infty\left(\frac{1}{2^{p-1}}\right)^n$ and this last is a geometric series which we know converges iff its quotient is between -1 and 1, i.e. iff

$\displaystyle \frac{1}{2^{p-1}}<1\Longleftrightarrow2^{p-1}>1\Longleftrightarrow p-1>0$ and we're done

Tonio