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So, I know how to show that the series $\sum_{n=0}^\infty (n^{1/n}-1)^n$ converge absolutely (I am not sure how this is related but is the first part of the problem).

Then they ask us to study if $\sum_{n=0}^\infty n^{1/n}-1$ converges. Other thing that I know is that $\sum_{n=0}^\infty n^{1/n}$ diverges. But then I get two bounds $(n^{1/n}-1)^n<n^{1/n}-1<n^{1/n}$ that don't help much...since the lower converge and the upper diverge.

I tried it in wolfram and it says that the series diverge by a comparison test. I tried comparing it with $n^{1/n}$ for $n\to \infty$ but I get $0$... which doesn't work either.

Any ideas on which serie can I use?

leplata
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3 Answers3

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HINT: For $n\ge 3$: $$n^{\frac{1}{n}} = e^{\frac{\ln n}{n}}$$ $$\ge 1+\frac{\ln n}{n} >1+ \frac{1}{n}.$$

Mike
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    $\forall n\ge3$. – PNDas Jun 08 '21 at 02:22
  • @PNDas good catch. Edited my answer accordingly – Mike Jun 08 '21 at 02:24
  • shouldn't this work also for n=1 and n=2? I mean $e^{x}\geq 1+x$ since for $x=0$ they are equal and then, for $x>0$, they are both positive but $e^{x}$ grows faster. – leplata Jun 08 '21 at 03:13
  • $\ln x < 1$ for $x <e$, so including $1,2$.. But for the purposes of the original exercise, it does not matter, showing that the inequality holds for $n \ge 3$ suffices. – Mike Jun 08 '21 at 03:23
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    nevermind... I realised that the $n\geq 3$ is for the last inequality. – leplata Jun 08 '21 at 03:23
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Hint $$n^\frac{1}{n}-1 > 2^\frac{1}{n}-1 \\ \lim_{n \to \infty} \frac{2^\frac{1}{n}-1}{\frac{1}{n}}= \ln(2) $$ the last limit following from the definition of the derivative of $2^x$ at $x=0$. Conclude from the second that $\sum_{n}2^\frac{1}{n}-1$ is divergent.

N. S.
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In this answer Bernoulli's Inequality is used to show that $$ \left(1+\frac1n\right)^{n+1} $$ is decreasing. Thus, for $n\ge1$, $$ \left(1+\frac1n\right)^{n+1}\le\left(1+\frac11\right)^{1+1}=4 $$ Therefore, if we substitute $n\mapsto n-1$ and take $n^\text{th}$ roots, we get that for $n\ge4$, $$ 1+\frac1{n-1}\le4^{1/n}\le n^{1/n} $$ Thus, $$ \sum_{n=4}^\infty\left(n^{1/n}-1\right)\ge\sum_{n=4}^\infty\frac1{n-1} $$ which diverges.

robjohn
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