# Does the series converge or diverge?

## $\infty$ sum_(n=1)1/(n^(1+1/n)

Jun 28, 2018

It diverges, since it is asymptotically equivalent to $\frac{1}{n}$

#### Explanation:

Let's use the comparison test. It goes like this: if you want to know if a series ${a}_{n}$ converges, and

$\setminus {\lim}_{n \setminus \to \setminus \infty} \setminus \frac{{a}_{n}}{{b}_{n}} = c$

where both ${a}_{n}$ and ${b}_{n}$ are sequences with positive terms and $c$ is finite, theny both ${a}_{n}$ and ${b}_{n}$ converge or diverge.

In this case, since

$\setminus {\lim}_{n \setminus \to \setminus \infty} \setminus \frac{1}{{n}^{1 + \setminus \frac{1}{n}}} = \setminus \frac{1}{n}$

we may try to use ${b}_{n} = \frac{1}{n}$ as comparison. Ideed, we have

$\setminus {\lim}_{n \setminus \to \setminus \infty} \setminus \frac{\setminus \frac{1}{{n}^{1 + \setminus \frac{1}{n}}}}{\frac{1}{n}} = \setminus {\lim}_{n \setminus \to \setminus \infty} \setminus \frac{{n}^{- 1 - \frac{1}{n}}}{{n}^{- 1}} = \setminus {\lim}_{n \setminus \to \setminus \infty} {n}^{\left(- 1 - \frac{1}{n}\right) - \left(- 1\right)}$

$= \setminus {\lim}_{n \setminus \to \setminus \infty} {n}^{- \frac{1}{n}} = 1$

So, the two series behave the same. Sine

$\setminus {\sum}_{n = 1}^{\setminus} \infty \frac{1}{n} = \setminus \infty$

then your series diverges as well.