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Pes 2012 Crack only reloaded CrackQ:

On the series $n^{\log_2n}$

It’s known that $$\sum_{n\ge 1} n^{\log_2n}=\infty\tag 1$$
(I saw it in this thread). Is the following sequence convergent too?
$$x_n:=n^{\log_2n}\tag 2$$
I tried to show that $x_n/n$ is monotonically decreasing, hence $\lim x_n/n$ exists and is finite. But I could not get a limit. Is it true? If so, how? If not, what is the right way to prove it?

A:

Yes, the given series is divergent. As Ron Gordon points out in a comment, we have that
$$n^{\log_2n} = n^{\frac{\log n}{\log 2}} \geq n^{\frac{1}{e}},$$
and the harmonic series diverges.
In fact, you can prove much more with just elementary properties of the logarithm:
$$n^{\log_2n} \geq n^{\log_2\log_2n} \geq n^{\log_2\log_2\log_2n} \geq \dots$$
If $\log_2n$ is prime for all sufficiently large $n$, then you get convergence. If not, then it isn’t.

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