How can we prove $\sum {\frac{1}{n^{1+1/n}}}$ is divergent?

Question

We know that $\sum \frac{1}{n^p}$ is convergent for $p>1$. However the series $\sum {\frac{1}{n^{1+1/n}}}$ is apparently divergent since $1+1/n$ tends to 1 as $n$ tends to infinity. But how to prove this? The root test fails expectedly and I haven't been able to find a smaller divergent series for comparison test.

Answer 1

Note that $n^{1/n} \lt 2$ for all $n$, for it is easy to show that $n \lt 2^n$. One can do this by induction, or by using the Binomial Theorem on $(1+1)^n$, or in several other ways.

It follows that $\dfrac{1}{n^{1+1/n}}\gt \dfrac{1}{2n}$.

Answer 2

The limit comparison test:

$$\frac{\frac{1}{n^{1+1/n}}}{\frac{1}{n}}=\frac{1}{\sqrt[n]n}\xrightarrow[n\to\infty]{}1$$

Thus, the series $\,\displaystyle{\sum_{n=1}^\infty\frac{1}{n\sqrt[n] n}\;\;,\;\;\sum_{n=1}^\infty\frac{1}{n}}\,$ converge/diverge together...

Answer 3

HINT

$1$. Prove that $\displaystyle \sum_n \dfrac1{n \log n}$ diverges using the fact that if we have a monotone decreasing sequence, then $\displaystyle \sum_{n=2}^{\infty} a_n$ converges iff $\displaystyle \sum_{n=2}^{\infty} 2^na_{2^n}$ converges

$2$. Prove that $\log n > n^{1/n}$ eventually and hence $$\dfrac1{n^{1/n}} > \dfrac1{\log n}$$

$3$. Conclude what you want.

Answer 4

By another way, using that $e^{-x}\ge 1-x$

$$n^{-1-\frac1n} =\frac1n \,e^{-\frac{\log n}n} \ge\frac1n-\frac{\log n}{n^2}$$

which implies divergence for the given series.