How do you test the series Sigma 1/(2+lnn) from n is [1,oo) for convergence?

1 Answer
Jan 30, 2018

${\sum}_{n = 1}^{\infty} \frac{1}{2 + \ln n}$

is divergent.

Explanation:

For $n \ge 2$ we have that $\ln n < n$, so that:

$\frac{1}{2 + \ln n} > \frac{1}{2 + n}$

From the limit comparison test we can also see that as:

${\lim}_{n \to \infty} \frac{\frac{1}{n + 2}}{\frac{1}{n}} = {\lim}_{n \to \infty} \frac{n}{n + 2} = 1$

and as:

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

is divergent, then also:

${\sum}_{n = 1}^{\infty} \frac{1}{n + 2}$

is divergent.

But then based on the direct comparison test also:

${\sum}_{n = 1}^{\infty} \frac{1}{2 + \ln n}$

is divergent.