# How do you test for convergence of Sigma n e^-n from n=[1,oo)?

Apr 1, 2017

The series:

${\sum}_{n = 1}^{\infty} n {e}^{- n}$

is convergent.

#### Explanation:

We can determine the convergence of the series:

${\sum}_{n = 1}^{\infty} n {e}^{- n}$

using the ratio test:

${\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid = {\lim}_{n \to \infty} \frac{\left(n + 1\right) {e}^{- \left(n + 1\right)}}{n {e}^{- n}}$

${\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid = {\lim}_{n \to \infty} \frac{n + 1}{n} \frac{{e}^{- n} {e}^{- 1}}{{e}^{- n}}$

${\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid = \frac{1}{e} {\lim}_{n \to \infty} \frac{n + 1}{n} = \frac{1}{e} < 1$

As the limit is less than $1$ the series is convergent.

Apr 1, 2017

See below.

#### Explanation:

${\sum}_{k = 0}^{m} {e}^{n x} = \frac{{e}^{\left(m + 1\right) x} - 1}{{e}^{x} - 1}$

${\sum}_{k = 0}^{m} n {e}^{n x} = \frac{d}{\mathrm{dx}} {\sum}_{k = 0}^{m} {e}^{n x} = \frac{{e}^{x} + {e}^{\left(2 + m\right) x} m - {e}^{\left(1 + m\right) x} \left(1 + m\right)}{{e}^{x} - 1} ^ 2$

and

${\sum}_{k = 0}^{m} n {e}^{- n x} = \frac{{e}^{-} x + {e}^{- \left(m + 2\right) x} m - {e}^{- \left(m + 1\right) x} \left(m + 1\right)}{{e}^{-} x - 1} ^ 2$

so, as we can observe, the series is convergent for $x \in {\mathbb{R}}^{+}$

because

${\lim}_{m \to \infty} \frac{{e}^{-} x + {e}^{- \left(m + 2\right) x} m - {e}^{- \left(m + 1\right) x} \left(m + 1\right)}{{e}^{-} x - 1} ^ 2 = - {e}^{x} / {\left({e}^{x} - 1\right)}^{2}$

and for $x = 1$ the limit is

$\frac{e}{e - 1} ^ 2$

Apr 1, 2017

The series is convergent by the ratio test.

#### Explanation:

${\sum}_{n = 1}^{\infty} n {e}^{- n}$

$= {\sum}_{n = 1}^{\infty} \frac{n}{e} ^ \left(n\right)$

Use the ratio test:
$\textcolor{b l u e}{\text{If "lim_(n->oo)|a_(n+1)/a_n|" where "a_n=n e^(-n) "is "< 1, "then the series is convergent,}}$
$\textcolor{b l u e}{\text{and if the limit is >1, the series diverges}}$

${\lim}_{n \to \infty} | \frac{\left(\frac{n + 1}{e} ^ \left(n + 1\right)\right)}{\left(\frac{n}{e} ^ n\right)} |$

(Dividing is the same as multiplying by reciprocal of denominator)
$= {\lim}_{n \to \infty} | \frac{\left(n + 1\right)}{\textcolor{red}{\left({e}^{n}\right)} \left(e\right)} \cdot \frac{\textcolor{red}{\left({e}^{n}\right)}}{n} |$

$= {\lim}_{n \to \infty} | \frac{\left(n + 1\right)}{\left(n\right) \left(e\right)} |$

$\textcolor{m a \ge n t a}{\text{Use the fact that "lim_(n->oo)(frac{n+1}{n})=1" to simplify}}$

$= {\lim}_{n \to \infty} | \frac{1}{e} |$

$= \frac{1}{e}$, which is less than 1.

Since the ratio test gives a value less than one, the series is convergent by the ratio test.