Determine whether this is convergent or divergent.
$$\sum _{n=1}^{\infty }\:\frac{\left(\ln n\right)^{10}}{n^{1.1}}$$
I tried comparing it with a harmonic sum to see if I can prove convergence, but I can't seem to find a proper answer. I really have a hard time dealing with sums that you need to use the comparison test with.
Best Answer
There exists $N$ such that for $n > N$, $$ \ln n < n^{.005}. $$ Then for $n > N$ we have $$ \frac{(\ln n)^{10}}{n^{1.1}} < \frac{n^{.05}}{n^{1.1}} = \frac{1}{n^{1.05}}. $$ Since $\sum_{n=1}^\infty \frac{1}{n^{1.05}}$ converges, this sum converges by the comparison test.