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

Question

Emilia Aguilar · Accepted Answer

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

#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#

Ezekiel Alexander · Answer

undefined The value of $ \frac{1}{n} \sum_{k=1}^n e^{k/n} $ is $ \frac{e-1}{e} $.