# Use comparison test to determine convergence of the following series?

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

Mar 27, 2018

${\sum}_{n - 1}^{\infty} \frac{n + 1}{n} ^ 2$ is divergence

#### Explanation:

Let ${a}_{n}$ = $\frac{n + 1}{n} ^ 2$ and ${b}_{n}$ = $\frac{1}{n}$ are sequence of positive real numbers.

such that

$L$ = lim_(n->+∞)((n+1)/n^2)/(1/n) = lim_(n->+∞)((n+1)/n) = $1$

Because

${\sum}_{n - 1}^{\infty} \left(\frac{1}{n}\right)$ is divergence,

so, by using limit ratio test (comparison test),

we said that ${\sum}_{n - 1}^{\infty} \frac{n + 1}{n} ^ 2$ is divergence.