Convergent sequence?

Question

Why does the $\lim \sum{n^{(1/n)}-1}$ diverge as $n\rightarrow \infty$? I suspected it would converge as $\lim {n^{(1/n)}}=1$ as $n\rightarrow \infty$ but computation show otherwise. So, now I am trying to show the partial sums are unbounded but didnt get far.

Answer 1

Hint: for $n\ge3$ we have $$n^{1/n}>e^{1/n}>1+\frac1n\ .$$

Answer 2

$$ n^{1/n}=1+a_n\quad \text{with}\quad \lim_{n\to\infty}a_n=0. $$ Taking logarithms $$ \frac{\log n}{n}=\log(1+a_n)\sim a_n\quad\text{as}\quad n\to\infty $$ and $\sum(\log n)/n$ diverges.

Answer 3

Check out the comparison test here, and find a series to compare with yours.

Answer 4

$n^{1/n} = e^{\ln(n)/n} - 1$ is positive and equivalent to the positive term $\ln(n)/n$ as $n\rightarrow +\infty$. (Just expand $\exp$ at the first order in $0$.) Not $\ln(n)/n$ is the general term of a Bertrand series which is in this case divergent. By equivalence, the series of general term $n^{1/n} = e^{\ln(n)/n} - 1$ is also divergent.