What's the range of $a$ when $\displaystyle \sum_{n=1}^\infty \frac{(\log_e n)^{2012}}{n^a} $ is convergent?

Question

What's the range of $a$ when $\displaystyle \sum_{n=1}^\infty \frac{(\log_e n)^{2012}}{n^a} $ is convergent?

Ratio test doesn't work. By root test I think a>0 is enough, but the answer is a>1.

Answer 1

PRIMER:

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1<x}\tag{P1}$$

for $x>0$. Now, using $\log(x^b)=b\log(x)$, then we see from $(P1)$ that for $b>0$,

$$\log(x)\le \frac{x^b-1}{b}<\frac{x^b}{b}\tag {P2}$$

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Note from $(P2)$, with $x=n$, that $\log(n)\le \frac{n^b-1}{b}<\frac{n^b}{b}$ for all $b>0$. Therefore,

$$\frac{\log^{2012}(n)}{n^a}<\frac{n^{2012b}}{bn^a} \tag 1$$

The inequality in $(1)$ is valid for all $b>0$ including those values of $b$ such that $a-2012b>1$. Therefore, taking $b<\frac{a-1}{2012}$, we see that the series converges by the "$p$ test" for $a>1$.

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Just as a numerical example, suppose $a=1.001$. Then, we can take $b<\frac{1}{2,012,000}$, say $b=\frac{1}{4,012,000}$, and thus

$$\frac{\log^{2012}(n)}{n^{1.001}}<\left(\frac{4,012,000}{n^{1.0005}}\right)$$

And the series $\sum_{n=1}^\infty\left(\frac{4,012,000}{n^{1.0005}}\right)$ converges.

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Now, if $a=1$ the series diverges. We can use the integral test. Simply write

$$\int_1^L \frac{\log^{2012}(x)}{x}\,dx=\frac{\log^{2013}(L)}{2013}\to \infty$$

The integral test shows that the series diverges for $a=1$

If $a<1$, then the series diverges by comparison with the harmonic series since for $a<1$

$$\frac{\log^{2012}(n)}{n^a}\ge \frac{1}{n}$$

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Putting it all together, we see that the series $\sum_{n=1}^\infty \frac{\log^{2012}(n)}{n^a}$ converges for all $a>1$ and diverges for all $a\le 1$.