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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!

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2 Answers 2

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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 $a\le1$

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  • $\begingroup$ 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. $\endgroup$
    – Ri-Li
    Nov 14, 2019 at 8:04
  • $\begingroup$ @Ri-Li We have indeed $$\frac{\left(\frac{\log n}n\right)^a}{\frac1{n^{\frac{1+a}2}}}=\frac{(\log n)^a}{n^{\frac{a-1}2}}\to 0$$ $\endgroup$
    – user
    Nov 14, 2019 at 8:06
  • $\begingroup$ Yeah, so true... $\endgroup$
    – Ri-Li
    Nov 14, 2019 at 8:08
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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 $a>1$.

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