Show that $\frac{1}{n^{\frac{n+1}{n}}}$ decreases How can one show that $\frac{1}{n^{\frac{n+1}{n}}}$ decreases?  I am using this to show that $\Sigma\frac{1}{n^{\frac{n+1}{n}}} $ diverges using Cauchy's condensation test.
If you know it decreases, then
$$ \varphi(n) \text{ is decreasing } \iff \varphi_{n+1} \leq \varphi_n$$

that is,

$$ (n+1)^{\frac{n+2}{n+1}} \geq n^{\frac{1+n}{n}}$$
can you take it from here? (hint: $\log\varphi_{n+1}\geq \log\varphi_n$)
If we look at the exponent in the expression: $\frac{n+1}{n}$, this simplifies to $1 + \frac{1}{n}$.

We clearly see that $\lim_{n\to\infty} \frac{1}{n} = 0$, which means the exponent is just $1 + 0 = 1$.

Then $\frac{1}{n^{\frac{n+1}{n}}}$ becomes $\frac{1}{n^{1}} = \frac{1}{n}$. But we already know what happens with $\frac{1}{n}$ when $n\to\infty$.

Thus giving you: $$\lim_{n\to\infty} \frac{1}{n^{\frac{n+1}{n}}} = 0$$

As the expression approaches $0$ when $n\to\infty$, it obviously decreases. Then use that fact in Cauchy's condensation test.