For which values of $a$ does the series converge?

For which values of $$a$$ does the series below converge? $$\sum_{n=1}^∞ \frac{(\ln n)^{2014}}{n^a}.$$ The answer is $$a > 1$$.

I do not have any idea how to do it. I have tried the ratio test and the integral test, but I still cannot figure it out. Can anyone help me with this? Thanks!

1 Answer

For $$\alpha \leq 1$$, comparison with the harmonic series shows the sum won't converge.

For $$\alpha > 1$$, it depends on how much analysis you've done, but if you can accept the following fact the proof is pretty easy:

$$\frac{\log(n)^{2014}}{n^{\beta}} \to 0 \quad \text{as}\ n \to \infty$$ for all $$\beta > 0$$. For convenience let $$\alpha = 1 + \beta$$ where $$\beta > 0$$. Then for sufficiently large $$n$$, $$\frac{\log(n)^{2014}}{n^{\beta/2}} \leq 1$$ thus we can compare with the sum of $$n^{-(1 + \beta/2)}$$ which converges. If you need further explanation on any part, please say.