# A_n := n^(1/n) - 1 yields a divergent series

## Homework Statement

I know that if $a_n := n^{1/n} - 1$, then $\Sigma a_n$ is divergent. I know this (by the integral test) because the integral of $2^{1/n} - 1$ from 1 to infinity is infinite. However, I want to avoid using non-elementary functions (here, the exponential integral) in my proof that this series is divergent.

Can anyone see a way of doing this?

2. Homework Equations /attempted solution
$lim\ sup_{n\to\infty} a_n^{1/n} = 1$, so the root test is inconclusive. Comparison is getting me nowhere. I'm thinking about seeing whether using the Taylor expansion of each term of the sequence shows what I want to show...

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I think the integral test is a good start. So we must prove that

$\int_1^{+\infty}{(2^{1/x}-1)dx}$

diverges.

Perhaps you can use the Taylor expansion of $2^{1/x}$ to find a good lower bound??

I think the integral test is a good start. So we must prove that

$\int_1^{+\infty}{(2^{1/x}-1)dx}$

diverges.

Perhaps you can use the Taylor expansion of $2^{1/x}$ to find a good lower bound??
Yes, thanks. As you'll see under number 2 above, that's exactly what I was considering doing.

n^(1/n) - 1 > 2^(1/n) - 1;
The definite integral of ( 2^(1/n) - 1 ) dn from 1 to infinity equals to
the definite integral of (2^x - 1)(-1/(x^2)) dx from 1 to 0 by making a substitution (x=1/n),
which, of course, equals to
the definite integral of ( (2^x )/(x^2) - 1/(x^2) )dx from 0 to 1.
Let f(x) = (2^x )/(x^2) - 1/(x^2), g(x)= - 1/(x^2) , then we have f(x) > g(x) for 0<x<1

Since we know the definite integral of g(x) dx from 0 to 1 equals to infinity,
we obtain that
The definite integral n^(1/n) - 1 dn from 1 to infinity also equals to infinity. [Finished]

Danny.
email: danny.s.deng.ds@gmail.com