sum_(k=1)^n e^(k/n) =sum_(k=0)^(n-1)e^(1/n)e^(k/n)
= sum_(k=0)^(n-1)e^(1/n)(e^(1/n))^k
By the geometric sum formula
sum_(k=0)^(n-1)ar^k = a(1-r^n)/(1-r)
we have
sum_(k=1)^n e^(k/n) = e^(1/n)(1-(e^(1/n))^n)/(1-e^(1/n))
= e^(1/n)(1-e)/(1-e^(1/n))
And so
1/nsum_(k=1)^n e^(k/n) = e^(1/n)(1-e)/(n(1-e^(1/n)))
To evaluate the above as n->oo, note that
e^(1/n)(1-e)/(n(1-e^(1/n))) = e^(1/n)(1-e)* 1/(n(1-e^(1/n)))
Thus, if
lim_(n->oo)e^(1/n)(1-e)
and
lim_(n->oo)1/(n(1-e^(1/n)))
both converge, then
lim_(n->oo)e^(1/n)(1-e)/(n(1-e^(1/n))) = lim_(n->oo)e^(1/n)(1-e)*lim_(n->oo)1/(n(1-e^(1/n)))
(*)
By direct substitution,
lim_(n->oo)e^(1/n)(1-e) = e^(1/oo)(1-e)
= e^0(1-e)
=1-e
Unfortunately, we cannot use direct substitution on
lim_(n->oo)1/(n(1-e^(1/n)))
as this gives us 0*oo in the denominator.
Instead, we will modify the expression so we can use l'Hopital's rule.
lim_(n->oo)1/(n(1-e^(1/n))) = lim_(n->oo)(1/n)/(1-e^(1/n)) which gives us the 0/0 form needed to apply l'Hopital's rule. Doing so, we have
lim_(n->oo)(1/n)/(1-e^(1/n)) = lim_(n->oo)(d/dx(1/n))/(d/dx(1-e^(1/n)))
= lim_(n->oo)(-1/n^2)/(e^(1/n)/n^2)
= lim_(n->oo) -1/e^(1/n)
We can now evaluate this by direct substitution.
lim_(n->oo) -1/e^(1/n) = -1/e^(1/oo)
= -1/e^0
= -1
Meaning we have lim_(n->oo)1/(n(1-e^(1/n))) = -1
So by our initial statement (*)
lim_(n->oo)e^(1/n)(1-e)/(n(1-e^(1/n))) = (1-e)*(-1) = e-1
