Question

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

Answer 1

`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

Answer 2

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.