# Consider $\sum_{n=1}^\infty(n^{1/n}-1)^a$. Find out all the values of $a$ for which this converges.

Consider $$\sum_{n=1}^\infty(n^{1/n}-1)^a$$ Find out all the values of $$a$$ for which this converges.

Let me include my try:

1. First observation, we know that $$\operatorname{log}x \leq x-1$$. Now putting $$x=n^{\frac1n}$$ we get $$\operatorname{log}n^{\frac1n} \leq n^{\frac1n}-1 \Rightarrow \frac1n \operatorname{log} n \leq n^{\frac1n}-1\Rightarrow \operatorname{log}n \leq n(n^{\frac1n}-1)\Rightarrow 1 \leq \operatorname{log}n \leq n(n^{\frac1n}-1)\text{ for n>3}\Rightarrow \frac1n \leq \frac {\operatorname{log}n} n\leq (n^{\frac1n}-1)\text{ for n>3}\Rightarrow (\frac1n)^a \leq (\frac {\operatorname{log}n} n)^a\leq (n^{\frac1n}-1)^a\text{ for n>3}$$

Now $$\sum_{n=1}^\infty (\frac1n)^a$$ diverges for all $$a \leq 1$$ so we get $$\sum_{n=1}^\infty(n^{\frac1n}-1)^a$$ diverges for all $$a \leq 1$$.

1. Second observation, $$\lim_{x \to 1} \frac{\operatorname{log}x}{ x-1}=1$$. So putting $$x=n^{\frac1n}$$ and observing that $$x \to 1$$ if $$n \to \infty$$, we get $$\lim_{n \to \infty} \frac{\operatorname{log}n^{\frac1n}}{ n^{\frac1n}-1}=1 \Rightarrow \lim_{n \to \infty} (\frac{\operatorname{log}n^{\frac1n}}{ n^{\frac1n}-1})^a=1$$.

So, $$\sum_{n=1}^\infty(n^{1/n}-1)^a$$ and $$\sum_{n=1}^\infty(\operatorname{log}n^{\frac1n})^a$$ converge and diverge simultaneously. So we might find a condition for what values of $$a$$, $$\sum_{n=1}^\infty(\operatorname{log}n^{\frac1n})^a$$ converges!

We have that

$$n^{1/n}=e^{\frac{\log n}n}\sim1+\frac{\log n}n$$

therefore

$$(n^{1/n}-1)^a \sim \left(\frac{\log n}n\right)^a$$

and the series converges by limit comparison test with $$\sum \frac1{n^{\frac{1+a}2}}$$ for any $$a>1$$ and diverges by limit comparison test with $$\sum \frac1{n}$$ for any $$a\le1$$

• Could you pls add the proof of "the series converges by limit comparison test with $\sum \frac1{n^{\frac{1+a}2}}$"? I have given the proof of the line above it. Nov 14, 2019 at 8:04
• @Ri-Li We have indeed $$\frac{\left(\frac{\log n}n\right)^a}{\frac1{n^{\frac{1+a}2}}}=\frac{(\log n)^a}{n^{\frac{a-1}2}}\to 0$$
– user
Nov 14, 2019 at 8:06
• Yeah, so true... Nov 14, 2019 at 8:08

We can use the integral test for $$\frac{\log^a n}{n^a}$$:

$$\int_1^\infty \left(\frac{\log x}{x}\right)^a dx = \int_0^\infty u^a e^{(1-a)u}du$$

According to the integrand, this integral diverges exactly when $$a\leq 1$$, but converges whenever $$a>1$$.