For which $p>0$ does $\sum_{n=3}^{\infty}\frac{\log(n)}{n^p}$ converge?

Question

For which $p>0$ does $\sum_{n=3}^{\infty}\frac{\log(n)}{n^p}$ converge? I tried all the criteria for series convergence I know, but I'm not getting any further with this exercise. I'm not asking someone to do my homework for me, but could somebody tell me what criteria I should try please or how to proceed?

Answer 1

Since $\log(x)/x^c \to 0$ for any $c > 0$, the sum converges if and only if $p > 1$.

Answer 2

What I would do here, is to use that $\log(n)$ is dominated by any $n^{\epsilon}$ for all $\epsilon>0$, provided $n$ large enough (to prove this you can use l'Hopital's rule by looking at the fraction of $n^{\epsilon}/\log n$ in the limit). Thus the tail of the summation is dominated by $n^{\epsilon - p}$ which converges if $\epsilon - p < -1$, or $p > 1+\epsilon$. Now, let $\epsilon$ tend to $0$ to get convergence for all $p>1$.

To see that it doesn't converge for $p\leq 1$, it is enough to note that $\log(n)\geq 1$. Thus $\log(n)/n^p \geq n^{-1}$ and hence, $$ \sum_{n=1}^{N} \frac{\log n}{n^p} \geq \sum_{n=1}^{N} \frac{1}{n} \stackrel{N\to\infty}{\longrightarrow} \infty. $$

I hope this helps and if you have any questions, feel free to comment!

Answer 3

Consider $3$ cases:

Case #1: $p=1$. From this: $log(n)>1 \implies \dfrac{log(n)}{n} > \dfrac{1}{n} $ you can see by comparison test (with harmonic series) that series diverges.

Case #2: $p>1$. You just use integration test and integrate this $\int_{1}^{+\infty} x^{-p} log(x)dx$ via integration by parts (set $u=log(x)$) and you will see that series converges (because the integral converges).

Case #3: $p<1$. You can easily deduce from previous cases that in this case sum diverges.

Answer 4

We can use the Cauchy condensation test

$$ 0 \ \leq\ \sum_{n=1}^{\infty} f(n)\ \leq\ \sum_{n=0}^{\infty} 2^{n}f(2^{n})\ \leq\ 2\sum_{n=1}^{\infty} f(n)$$

that is

$$\sum_{n=3}^{\infty}\frac{\log(n)}{n^p}\le \sum_{n=3}^{\infty}\frac{2^n\log(2^n)}{2^{np}}=\sum_{n=3}^{\infty}\frac{n\log(2)}{2^{n(p-1)}}\le 2\sum_{n=3}^{\infty}\frac{\log(n)}{n^p}$$

Answer 5

As already mentioned elsewhere, this is (apart from the sign and the missing contribution from n=2) the first derivative of the Riemann Zeta function at p.