#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#
