How could I show that the following sequence converges?
$$\sum_{n = 1}^{\infty} \frac{\sqrt{n} \log n}{n^2 + 3n + 1}$$
I tried the ratio and nth-root tests and both were inconclusive. I was thinking there might be a way to use the limit comparison test, but I'm not sure. Any hints?
Hint:
$$\sum_{n = 1}^{\infty} \frac{\sqrt{n} \log n}{n^2 + 3n + 1} < \sum_{n = 1}^{\infty} \frac{ \log n}{n^{3/2}}$$
Then by integral test, since $\int_{1}^{\infty}\frac{ \log n}{n^{3/2}}=4$ (converges), so the given series converges.
You can also try the limit comparison test with $\;\frac{\log n}{n^{3/2}} \;$? :
$$\frac{\frac{\sqrt n\log n}{n^2+3n+1}}{\frac{\log n}{n^{3/2}}}=\frac{n^2}{n^2+3n+1}\xrightarrow[n\to\infty]{}1$$
Finally, note that the series $\;\sum\frac{\log n}{n^{3/2}}\;$ converges since, for example:
$$\begin{align*}\text{Comparison Test:}&\;\frac{\log n}{n^{3/2}}\le\frac{n^{1/4}}{n^{3/2}}=\frac1{n^{5/4}}\\{}\\ \text{Condensation Test:}&\;\;\frac{2^n\log2^n}{2^{3n/2}}=\log 2\frac n{2^{n/2}}\end{align*}$$
and the last one is convergent (for example, $\;n-$ th root test)
