convergence or divergence of $\sum_{n=0}^∞ \frac{(\ln(n)+5)}{n^2}$

Question

Determine whether the following series is convergent or divergent

$$\sum_{n=0}^∞ \frac{(\ln(n)+5)}{n^2}$$

I tried using comparison test and p-test: $$\frac{\ln(n)+5}{n^2}$$<=$$\frac{1}{n^2}$$

And since 1/n^2 is convergent by p-test, the series is convergent. But i am not sure if my solution is correct especially the part on determining the upper bound of $$\frac{\ln(n)+5}{n^2}$$

Answer 1

Note that$$\frac{\ln(n)+5}{n^2}\leqslant\frac1{n^{3/2}}\iff\ln(n)+5\leqslant\sqrt n,$$and this last statement holds, if $n\gg1$. So, since the series $\sum_{n=1}^\infty\frac1{n^{3/2}}$ converges, so does your series.

Answer 2

Hint: apply Cauchy's condensation test on the general term of the series.

$$2^k \cdot \frac{\ln(2^k) + 5}{(2^k)^2} = \frac{k\ln2 + 5}{2^{k}}$$

It's clear that the above series converge.

Answer 3

You can't use this comparison, because $$\lim\frac{\frac{\ln n+5}{n^2}}{\frac{1}{n^2}}=\infty$$ (You would have to have finite nonzero limit to get what you want from this test)

The fastest test to apply here, is the condensation test: $$ \sum 2^n\frac{\ln (2^n)+5}{(2^n)^2}=\sum\frac{n\ln 2+5}{2^n}<\infty $$ ($\sum\frac{n}{2^n}$ converges by Cauchy: $\sqrt[n]{\frac{n}{2^n}}\to\frac{1}{2}<1$) so your series converges by condensation test.