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

2 Answers
George C.
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

bp
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.

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