# Convergence of $\sum _{n=1}^{\infty }\:\frac{\left(\ln n\right)^{10}}{n^{1.1}}$

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.

• We don't really want to start at $n=0$. Do a Limit Comparison with $\sum_1^\infty \frac{1}{n^{1.05}}$. – André Nicolas Nov 11 '15 at 20:38
• You're right, should start at 1. If I do a limit comparison with the series you just mentioned wouldn't that yield me infinity? – MikhaelM Nov 11 '15 at 20:41
• In the long run the terms of your series are less than the terms of the series I proposed, That series converges, so your series converges. – André Nicolas Nov 11 '15 at 20:44
• How are they less? Aren't the terms in my series always going to be $\gt$ the terms in the proposed series? – MikhaelM Nov 11 '15 at 20:46
• In the long run $n^{0.05}$ is greater than any power of a logarithm. – André Nicolas Nov 11 '15 at 20:48

## 2 Answers

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.

This is a series with positive terms. Furthermore, for any $\alpha>0$, $(\ln n)^\alpha=o(n^{0.05})$, say. Hence $$\frac{\ln^\alpha n}{n^{1.1}}=o\biggl(\frac1{n^{1.05}}\biggl),$$ which converges.

• Not familiar with the $o$ notation(we haven't studied it yet) so I don't think I should be using that :( – MikhaelM Nov 11 '15 at 20:42
• Then you can prove the function $\dfrac{\ln^{10} x}{x^{0.05}}$ is bounded on $[1,+\infty)$. – Bernard Nov 11 '15 at 20:53
• @MikhaelM $f$ being little-$o$ of $g$ (written as $f=o(g)$) is simply a convenient way of saying that the function on the right "dominates" the function on the left. I.E. that $f=o(g)\Leftrightarrow \lim\limits_{n\to\infty}\frac{f(n)}{g(n)}=0 \Leftrightarrow f\prec g$ – JMoravitz Nov 11 '15 at 20:53