Test the convergence of $\sum_{n=1}^\infty \frac{n^{1/n} - 1}{n}$ [duplicate]

Question

Test the convergence of $$\sum_{n=1}^\infty \frac{n^{1/n} - 1}{n}$$

My Attempt: Using the root or ratio test would be too inconvenient here. Looking at the denominator, I used the Cauchy Condensation Test, i.e. if the condensed sequence $\sum_{k=1}^{\infty} 2^k a_{2^k}$ converges, then $\sum a_n$ converges. By this test, we have $$\sum_{k=1}^{\infty} 2^{k} \frac{(2^k)^{\frac{1}{2^k}} - 1}{2^k} = \sum_{k=1}^{\infty} 2^{\frac{k}{2^k}} - 1$$

Consider the term $2^{\frac{k}{2^k}} - 1$. If we rewrite 2 as a binomial $1+1$, we get $$(1+1)^{\frac{k}{2^k}} - 1 \lt \frac{k}{2^k}$$ Summing both sides, $$\sum_{k=1}^{\infty} 2^{\frac{k}{2^k}} - 1 < \sum_{k=1}^{\infty} \frac{k}{2^k}$$ For the latter sum, it is trivial to show that $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \frac{1}{2} < 1$, and the series converges by ratio test. Hence $\sum_{k=1}^{\infty} 2^{\frac{k}{2^k}} - 1$ converges and therefore, $\sum_{n=1}^\infty \frac{n^{1/n} - 1}{n}$ is convergent. $\; \blacksquare$

Is this proof valid? Please point out the errors and suggest improvements as well. I think there may be a less complicated way to solve this, but I'm not able to put my finger on it.

Answer 1

Another way: $\frac {e^{x}-1} x \to 1$ as $x \to 0$ so $e^{x}-1 <2x$ for $x>0$ close enough to $0$. So for $n$ sufficiently large we get $n^{1/n}-1 =e^{\frac1 n \ln n} -1< \frac {2 \ln n} n$. Now $\ln n=\ 2 \ln (n^{\frac 1 2})<2n^{1/2}$, Now you can compare the given series with $\sum \frac 1 {n^{3/2}}$ which is convergent.