# Does $\sum\limits_{n=1}^\infty \frac{1}{n^p}, 1 < p < 2$ converge?

I know $\sum\limits_{n=1}^\infty \frac{1}{n^p}$ diverges when $p \le 1$ and converges when $p \ge 2$. How about $1 < p < 2$?

• Converges, Integral Test. Commented Mar 12, 2016 at 2:11
• It does. You also have the slightly better statement: $\sum_{n=2}^{\infty} \frac{1}{n^a(\ln n)^b}$ converges iff $a > 1$ or ($a=1$ and $b>1$). (Such series are called Bertrand Series.) Commented Mar 12, 2016 at 2:14

By the Integral Test (note that the terms are monotone decreasing and positive), we may consider the integral

$$\int_1 ^ \infty \frac{1}{x^p} dx$$

Since $p > 1$, i.e. $p - 1 > 0$, we can evaluate the integral as

$$\left[\frac{1}{(1-p)x^{p-1}}\right]_1 ^ \infty = - \frac{1}{1-p} < \infty$$

The $p$-series converge for any $p > 1$. Observe that $$\sum_{n=1}^\infty 2^n \frac{1}{2^{np}} = \sum_{n=1}^\infty 2^{(1-p)n}$$ By the Cauchy condensation test, $\sum_{n=1}^\infty n^{-p}$ is bounded by the above sum, and hence coverges.

Here's another cool test, the Cauchy Condensation Test:

Let $(a_n)$ be a nonincreasing sequence of positive numbers. Then $\sum_{n=1}^{\infty} a_n$ converges if and only if $\sum_{n=1}^\infty 2^n a_{2^n}$ converges.

This is essentially a discrete change of variables (or substitution) formula, and as change of variables in integrals is a powerful tool, so is this test. The proof is very simple and follows from the monoonicity of the sequence. For example:

Harmonic series $a_n =n^{-p}$. Then $a_{2^n} =2^{-np}$, and

$$\sum_{n=1}^\infty 2^n a_{2^n} = \sum_{n=1}^\infty 2^{(1-p)n},$$

a geometric series, which converges if and only if $p>1$.

Another example. $a_n = 1/(n \ln^\alpha n)$. Then

$a_{2^n} = \frac{1}{2^n n^{\alpha} (\ln 2)^{\alpha}}$. Therefore

$\sum_{n=2}^{\infty}2^n a_{2^n} = \frac{1}{(\ln 2)^{\alpha}} \sum_{n=1}^\infty n^{-\alpha}$.

By last example this converges if and only if $\alpha>1$.

You can continue and do more and more complex examples. Try $a_n = \frac{1} {n (\ln n) (\ln\ln n ) \dots (\ln \ln \ln \dots \ln n)^{\beta}}$, when the number of iterations of $\ln$ is fixed and equal to $k$.

You already have two standard ways. Here is another, using comparison:

For $p>1$, we have $$\lim_{n\to+\infty}\frac{\frac{1}{n^{p-1}}-\frac{1}{(n+1)^{p-1}}}{\frac{1}{n^p}}= p-1\neq 0.$$ Now, the positive series $$\sum_{n=1}^{+\infty} \Bigl(\frac{1}{n^{p-1}}-\frac{1}{(n+1)^{p-1}}\Bigr)$$ converges (it is a telescoping series with terms tending to $0$, the sum is actually equal to $1$). Hence, by comparison $$\sum_{n=1}^{+\infty}\frac{1}{n^p}$$ also converges.