# Proving that the harmonic p series converges for p>1 and diverges for p<=1

Can someone please check if I have done this correctly? The harmonic p-series:

$$\sum_{n=1}^\infty \frac{1}{n^p}$$ $$let$$ $$f(n)=\frac{1}{n^p}$$

$$f(x)=\frac{1}{x^p}$$

Since f(x) is a positive, decreasing, continuous function, applying the integral test:

$$\int_0^\infty \frac{1}{x^p} = \int_0^\infty x^{-p} =\frac{1}{1-p} \lim_{h\to \infty} (h^{1-p} - 1) = \frac{1}{1-p} \lim_{h\to \infty} \frac{1}{h^{p-1}}$$ For this integral to converge (and the harmonic p-series) to converge: $lim_{h\to \infty} \frac{1}{h^{p-1}}$ must be be finite. For this to be the case, $p-1>0$ <--Should this be p-1>=0????-->

Therefore for convergence of the p-series, $p>1$, and for divergence, $p<=1$ by the integral test.

## 1 Answer

You're almost correct but take care with the limits of the integral:

the given series has the same nature (being convergent or divergent) as the integral

$$\int_{\color{red}{\pmb1}}^\infty\frac{dx}{x^p}$$ which's convergent if and only if $p>1$.

• You're welcome. – user63181 May 22 '14 at 15:43