Using the integral test, how do you show whether # (1 + (1/x))^x# diverges or converges?

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Answer 1 · 2015-12-23T21:30:40

it converge

By the way you can process like this

#(1+1/x)^x=e^(xln(1+1/x)#

#1/(x+1)<=ln(1+1/x)<=1/x#

#x/(x+1)<=xln(1+1/x)<=1#

#1/(1+1/x)<=xln(1+1/x)<=1#

Take the limit of #1/(1+1/x) # at #-oo# and #oo# for both case it's 1

by the squeeze theorem you can say that #lim x-> oo# #xln(1+1/x) = 1#
and #lim x-> -oo# #xln(1+1/x) = 1#

So #(1+1/x)^x# converge