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$$\sum_{n=1}^{\infty} \frac{1}{n \cdot n^{\frac{1}{n}}}$$

Is this infinite sum convergent?

There is a hint that using relation between $\ln n$ and $n^{\frac{1}{n}}$, but I don't know what the relation is.

cokecokecoke
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6 Answers6

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For $n > e^2$, $\ln n \ge n^{\frac1n}$. Indeed: $\ln^n n \ge 2^n \ge n$. Thus: $$\frac1{n \cdot n^{\frac1n}} \ge \frac{1}{n \ln n}$$

We know that $\sum \frac1{n \ln n} = \infty$ (integral test). Thus, by comparison test, the series diverges to $\infty$.

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    Even worse ... $n^{1/n}=O(1)$ as $n\to \infty$. Why bother using the integral test? – Mark Viola Apr 23 '17 at 16:30
  • @Dr.MV I avoid equivalents whenever I could. –  Apr 23 '17 at 16:32
  • Why would you avoid equivalents? They are perfectly rigorous. And certainly, we can transform this one into an inequality. -Mark – Mark Viola Apr 23 '17 at 16:37
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    It's easier to use $2>n^{1/n}$. – Akiva Weinberger Apr 23 '17 at 20:08
  • @AkivaWeinberger I misread OP's post thinking that they should use a relation between $\ln n$ and $n^{1/n}$. The whole thing is completely useless though, of course using something like $2 > n^{1/n}$ is much more sensible. –  Apr 23 '17 at 20:12
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HINT:

$$n^{1/n}=e^{\frac{1}{n}\log(n)}\to 1\,\,\text{as}\,\,n\to \infty$$

Mark Viola
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The series $\sum a_n$ is not convergent by the limit comparison test with $b_{n}=\frac{1}{n}$, and the corresponding harmonic series.

Another way to see this is that there exists some $N$ such that for all $n>N$ we have $n n^{1/n} < n \ln n$. This shows,

$$\sum_{n=N}^{\infty} \frac{1}{n n^{1/n}} > \sum_{n=N}^{\infty} \frac{1}{n \ln n}$$

The last series is well known to diverge by the integral test.

Ahmed S. Attaalla
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Try the limit comparison test. If $a_n=1/(n\cdot n^{(1/n)})$ and $b_n=1/n$ then limit of $b_n/a_n$ is $1$ so one gets your series converges or diverges iff the simpler series $1/n$ does.

coffeemath
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1

Another way to approach this:

Let $\displaystyle f(n) = n^{1/n} = e^{{\Large\frac{\ln(n)}{n}}}$. Then $\displaystyle f'(n) = e^{\Large\frac{\ln(n)}{n}} \left( \frac{1 - \ln(n)}{n^2} \right)$.

So $f'(n) = 0 $ for $n= e$, $f'(n) > 0$ for $n < e$ en $f'(n) < 0$ for $n > e$.

So there is a maximum for $n = e$ which is $e^{1/e} < e$, so $f(n)<e$ for all positive integers $n$.

This allows you to use the comparison test with $\sum_{n=1}^{\infty}\dfrac{1}{ne}$.

wythagoras
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Here is an easier proof:

For large $n$, we have $$n < 2^n \implies n^{\frac{1}n} < 2 \implies n^{1 + 1/n} < 2n \implies \frac{1}{n^{1+1/n}} > \frac{1}{2n}.$$ The result follows from the comparison test.

Sandeep Silwal
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