Question

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

Answer 1

The series:

`sum_(n=1)^oo n e^(-n)`

is convergent.

Explanation:

We can determine the convergence of the series:

`sum_(n=1)^oo n e^(-n)`

using the ratio test:

`lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) ((n+1)e^(-(n+1)))/(n e^(-n))`

`lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) (n+1)/n (e^(-n) e^(-1))/(e^(-n))`

`lim_(n->oo) abs (a_(n+1)/a_n) = 1/e lim_(n->oo) (n+1)/n = 1/e < 1`

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

Answer 2

See below.

Explanation:

`sum_(k=0)^m e^(nx) = (e^((m+1)x)-1)/(e^x-1)`

`sum_(k=0)^m n e^(nx) = d/(dx)sum_(k=0)^m e^(nx) = (e^x + e^((2 + m) x) m - e^((1 + m) x) (1 + m))/(e^x-1)^2`

and

`sum_(k=0)^m n e^(-nx) = (e^-x + e^(-(m+2) x) m - e^(-(m+1) x) (m+1))/(e^-x-1)^2`

so, as we can observe, the series is convergent for `x in RR^+`

because

`lim_(m->oo) (e^-x + e^(-(m+2) x) m - e^(-(m+1) x) (m+1))/(e^-x-1)^2=-e^x/(e^x-1)^2`

and for `x = 1` the limit is

`e/(e-1)^2`

Answer 3

The series is convergent by the ratio test.

Explanation:

`sum_(n=1)^(oo)n e^(-n)`

`=sum_(n=1)^(oo)n/ e^(n)`

Use the ratio test:
`color(blue)("If "lim_(n->oo)|a_(n+1)/a_n|" where "a_n=n e^(-n) "is "< 1, "then the series is convergent," )`
`color(blue)("and if the limit is >1, the series diverges")`

`lim_(n->oo)|frac{((n+1)/e^(n+1))}{(n/e^n)}|`

(Dividing is the same as multiplying by reciprocal of denominator)
`=lim_(n->oo)|frac{(n+1)}{color(red)((e^n))(e)}*frac{color(red)((e^n))}{n}|`

`=lim_(n->oo)|frac{(n+1)}{(n) (e)}|`

`color(magenta)("Use the fact that "lim_(n->oo)(frac{n+1}{n})=1" to simplify")`

`=lim_(n->oo)|1/e|`

`=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.