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

Question

Emilia Aguilar · Accepted Answer

#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#

Evelyn Agard · Answer

undefined To determine whether the infinite sequence $a_n = \left(1 + \frac{1}{n}\right)^n$ converges or diverges, we can analyze its behavior as $n$ approaches infinity. This sequence is a form of the limit of $(1 + \frac{1}{n})^n$ as $n$ approaches infinity, known as Euler's number, denoted by $e$. If $e$, approximately equal to 2.71828, is used to approximate the expression, it gives a clearer understanding of the sequence's behavior. As $n$ approaches infinity, the sequence $a_n$ converges to $e$, meaning it has a finite limit, rather than growing infinitely large or oscillating. Therefore, the infinite sequence $a_n = \left(1 + \frac{1}{n}\right)^n$ converges to Euler's number, $e$.