# how to prove $\sum {\frac{1}{n^{1+1/n}}}$ is divergent

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. Can someone help? Thanks!

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

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

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

• Thanks a lot for the swift reply!! The limit comparison test comes in handy! – Jason Ng Mar 4 '13 at 5:06

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.

• Thanks for answering so fast! Is there a quick way to explain $\log(n)>n^{1/n}$? Thanks! – Jason Ng Mar 4 '13 at 5:23
• @JasonNg $\log(n) \to \infty$ but $n^{1/n} \to 1$. For large enough $n$, then, we know $\log(n) > 2$ (say) but $n^{1/n}$ is close to $1$. – guy Mar 4 '13 at 5:58
• Thanks a lot! Should have thought of this! – Jason Ng Mar 4 '13 at 6:05