Question

How do you find the radius of convergence `Sigma (1+1/n)^-nx^n` from `n=[1,oo)`?

Answer 1

The radius of convergence of this series is `R=1`

Explanation:

We can apply the ratio test:

`a_n = (1-1/n)^-nx^n`

`abs (a_(n+1)/a_n) = abs( ( (1+1/(n+1))^-(n+1) x^(n+1)) / ((1+1/n)^-n x^n)) = ( (1+1/(n+1))^-(n+1) ) / ((1+1/n)^-n ) abs(x) = ( (1+1/n)^n ) / ((1+1/(n+1))^(n+1)) abs(x)`

As:

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

we have:

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

so the series is absolutely convergent for `abs(x) < 1`