Is sum_(n=0)^oo (1-1/n)^n convergent or divergent ?

2 Answers
Oct 4, 2015

sum_(n=0)^oo (1-1/n)^n is divergent.

Explanation:

First note that lim_(n->oo) (1+x/n)^n = e^x

We can see this using the binomial theorem (though I'm not sure this is rigorous)...

(1+x/n)^n = ((n),(0)) + ((n),(1))x/n + ((n),(2))x^2/(n^2) + ...

=1 + n/n x/(1!) + (n(n-1))/(n^2) x^2/(2!) + (n(n-1)(n-2))/(n^3) x^3/(3!) +...

=1 + x/(1!) + (n^2+O(n))/(n^2) x^2/(2!) + (n^3+O(n^2))/(n^3) x^3/(3!) +...

=1 + x/(1!) + (1+(O(n))/(n^2)) x^2/(2!) + (1+(O(n^2))/(n^3)) x^3/(3!) +...

-> sum_(k=0)^oo x^k/(k!) as n->oo

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

So EE N in NN : AA n >= N, abs((1-1/n)^n - e^(-1)) < e^(-1)/2

If abs((1-1/n)^n - e^(-1)) < e^(-1)/2, then (1-1/n)^n > e^(-1)/2

So sum_(n=0)^oo (1-1/n)^n > sum_(n=0)^(N-1) (1-1/n)^n + sum_(n=N)^oo e^-1/2 diverges

Oct 5, 2015

The series is divergent.

Explanation:

Apply the divergence test which says that if lim n-> oo, nth term a_n is !=0, then series diverges. In the present case, lim n->oo (1-1/n)^n = 1/e, hence the series would diverge.