# How do you test the series Sigma n^-n from n is [1,oo) for convergence?

Jun 4, 2017

Using the ratio test

#### Explanation:

The ratio test finds the ratio between terms ${u}_{k} \mathmr{and} {u}_{k + 1}$ as $k$ tends to infinity, and if $\left\mid {u}_{k + 1} / {u}_{k} \right\mid < 1 \left(k \to \infty\right)$, then the series is found to be convergent, as it means that terms are getting progressively smaller and tending towards 0.

Hence, in order to test $\sum {n}^{-} n , n \in \left[1 , \infty\right)$ for convergence, we find the ratio $\left\mid {u}_{k + 1} / {u}_{k} \right\mid \left(k \to \infty\right)$:

$k \to \infty \left\mid {u}_{k + 1} / {u}_{k} \right\mid$
$= k \to \infty \left\mid {\left(k + 1\right)}^{-} \frac{k + 1}{k} ^ - k \right\mid$
$= k \to \infty \left\mid {k}^{k} / {\left(k + 1\right)}^{k + 1} \right\mid$
$= k \to \infty \left\mid {k}^{k} / {\left(k + 1\right)}^{k} \cdot \frac{1}{k + 1} \right\mid$
$= k \to \infty \left\mid {\left(\frac{k}{k + 1}\right)}^{k} \cdot \frac{1}{k + 1} \right\mid$
$= k \to \infty \left\mid {\left(\frac{1}{1 + \frac{1}{k}}\right)}^{k} \cdot \frac{1}{k + 1} \right\mid$
$\because k \to \infty \frac{1}{k} = 0$
$\therefore k \to \infty \left\mid {\left(\frac{1}{1 + \frac{1}{k}}\right)}^{k} \right\mid = k \to \infty \left\mid {\left(\frac{1}{1 + 0}\right)}^{k} \right\mid = \left\mid 1 \right\mid = 1$
$\therefore k \to \infty \left\mid {\left(\frac{1}{1 + \frac{1}{k}}\right)}^{k} \cdot \frac{1}{k + 1} \right\mid = k \to \infty \left\mid 1 \cdot \frac{1}{k + 1} \right\mid$

$= k \to \infty \left\mid \frac{1}{k + 1} \right\mid = 0$

$\because k \to \infty \left\mid {u}_{k + 1} / {u}_{k} \right\mid = 0$
i.e. $k \to \infty \left\mid {u}_{k + 1} / {u}_{k} \right\mid < 1$
$\therefore$ By the ratio test, the series $\sum {n}^{-} n$ converges from $n \in \left[1 , \infty\right)$