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

1 Answer
Andrea S.
Jan 19, 2017

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

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