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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!

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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.

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