How do you determine whether the infinite sequence #a_n=(1+1/n)^n# converges or diverges?

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Answer 1 · 2014-09-22T04:17:28

#lim_{n to infty}a_n=e# (converges)

Let us look at some details.

By rewriting,

#a_n=e^{ln(1+1/n)^n}=e^{nln(1+1/n)}=e^{[ln(1+1/n)]/{1/n}}#

Now, let us evaluate the limit.

#lim_{n to infty}a_n=e^{[ln(1+1/n)]/{1/n}} =e^{lim_{n to infty}{ln(1+1/n)]/{1/n}#

by l'Hopital's Rule,

#=e^{lim_{n to infty}{{-1/n^2}/{1+1/n}]/{-1/n^2}#

by cancelling out #-1/n^2#,

#=e^{lim_{n to infty}1/{1+1/n}}=e^{1/{1+0}}=e#