Using the definition of a convergent sequence, prove limn→∞e(n+1n)=e (n+1n is the exponent of e) Don't use any theorems about convergent sequences

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Efan Halliday

Answered question

2021-08-17

Using the definition of a convergent sequence, prove
$lim_{n \to \infty} e^{(\frac{n + 1}{n})} = e$
$(\frac{n + 1}{n} is the exponent of e)$
Don't use any theorems about convergent sequences

Answer & Explanation

mhalmantus

Skilled2021-08-18Added 105 answers

$lim_{n \to \infty} a_{n} = a$ convergent if $\forall ϵ > 0$
There exist $a_{n}, N_{0} \in N$ such that
$| a_{n} - a | < ϵ$ when $n > N_{0}$
Choose $ϵ > 0$ such that
$| e^{(\frac{n + 1}{n})} - e | < ϵ$
$| e^{\frac{n + 1}{n} - 1} - 1 | < \frac{ϵ}{e}$
$| e^{\frac{1}{n}} - 1 | < \frac{ϵ}{e}$
$e^{\frac{1}{n}} - 1 < | e^{\frac{1}{n}} - 1 | < \frac{ϵ}{e}$
$e^{\frac{1}{n}} < \frac{ϵ}{e} + 1$
$\frac{1}{n} < \ln (\frac{ϵ}{e} + 1)$
$n > \frac{1}{\ln (\frac{ϵ}{e} + 1)}$
$N_{0} > \frac{1}{\ln (\frac{ϵ}{e} + 1)}$
So $\forall ϵ > 0$ we can find and $N_{0} \in N$ such that
$| e^{\frac{n + 1}{n}} - e | < ϵ$ when $n > N_{0}$
Hence $lim_{n \to \infty} e^{(\frac{n + 1}{n})} = e$

Jeffrey Jordon

Expert2021-10-23Added 2605 answers

Answer is given below (on video)

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