What is the value of #1/n sum_{k=1}^n e^{k/n}# ?

1 Answer
Dec 19, 2015

Answer:

#e-1# (after taking the limit as #n->oo#)

Explanation:

#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#