Is the infinite sum convergent? $$\sum_{n=1}^{\infty} \frac{1}{n \cdot n^{\frac{1}{n}}}$$
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.
 A: 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 $\infty$.
A: HINT:
$$n^{1/n}=e^{\frac{1}{n}\log(n)}\to 1\,\,\text{as}\,\,n\to \infty$$
A: 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.
A: 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.
A: 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 $\sum_{n=1}^{\infty}\dfrac{1}{ne}$.
A: 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. 
