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If I set N=$2\over n$, the equation becomes $(1+ N)^{2\over N}$, which I can take the natural log of and work down until I reach $e^2$. However, my teacher wants me to use subsequences, starting with $x_n$ = $(1+ {1\over n})^n = e$, to prove this and I am stuck.

Martin Sleziak
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p3ngu1n
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  • Related: http://math.stackexchange.com/questions/358830/about-lim-left1-frac-xn-rightn – Martin Sleziak Dec 17 '15 at 06:12
  • As a generalization, you might want to show that $\displaystyle \lim_{n\to\infty}\left(1+\frac{q}{n}\right)^n = e^q \text{ where }q\in\mathbb{R}$. – Galc127 Dec 17 '15 at 07:06

3 Answers3

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You could rewrite this as $$\lim_{n \to \infty}\left(1+\frac{1}{n/2}\right)^n = \lim_{n \to \infty}\left(\left(1+\frac{1}{n/2}\right)^{n/2}\right)^2 = \lim_{n\to\infty} (x_n)^2$$ where $x_n=\left(1+\frac{1}{n/2}\right)^{n/2}$. Now as $n \to \infty$, $\frac{n}{2} \to \infty$ as well. Hence $\lim_{n\to\infty} x_n = e$. Thus, bringing the limit inside the square (which we can do since squaring is continuous): $$\lim_{n \to \infty}\left(1+\frac{1}{n/2}\right)^n = \lim_{n \to \infty}(x_n)^2= \left(\lim_{n\to\infty} x_n \right)^2=e^2.$$ This doesn't "use subsequences," but it makes use of the sequence whose limit you already know, if that was your teacher's intention.

kccu
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Let y=(1+2/n)^n

ln⁡y=ln⁡〖(1+2/n)^n 〗

lim┬(n→∞)⁡ln⁡〖y= lim┬(n→∞)⁡ln⁡〖(1+2/n)^n 〗=ln⁡lim┬(n→∞)⁡y

〖lim┬(n→∞) n〗⁡ln⁡(1+2/n) =∞×0;hence undefined.

Proceed by using L’Hopitals Rule:

〖lim┬(n→∞) n〗⁡ln⁡(1+2/n) =lim┬(n→∞)⁡〖ln⁡(1+2/n)/(1/n)〗 =lim┬(n→∞)⁡〖((-2)/(n^2 (1+2/n) ))/((-1)/n^2 )〗 =lim┬(n→∞)⁡〖2/((1+2/n) )〗→2

Hence ln⁡lim┬(n→∞)⁡y =2=ln⁡〖e^2 〗. Hence lim┬(n→∞)⁡〖(1+2/n)^n 〗=e^2

Tosh
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  • I'm so sorry, I did not see the bit on subsequences.For subsequences, try using binomial theorem. I shall show this in another post – Tosh Dec 16 '15 at 07:26
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Let $a_n$ =$(1+2/n)^n$ and $a_{n+1}$=$1+$($\frac{2}{n+1})^{n+1} $

Proceed by their binomial expansion.

$a_n$=$1+n$($\frac{2}{n})$+n($\frac{n-1}{2!})$ $(\frac{2}{n})^{2}$+⋯+$\frac{n(n-1)(n-2)…}{n!}$ ($\frac{2}{n})^{n}$

$a_n=$1+2+($\frac{1}{2!})$ $2^2$(1-$\frac{1}{n})$+($\frac{1}{3!})$ $2^3$(1-$\frac{1}{n})$(1-$\frac{2}{n})$+⋯+($\frac{1}{n!})$ $2^n$(1-$\frac{1}{n})$(1-$\frac{2}{n})$...(1-$\frac{n-1}{n})$

$a_{n+1}=$1+2+($\frac{1}{2!})$ $2^2$(1-$\frac{1}{n+1})$+($\frac{1}{3!})$ $2^3$(1-$\frac{1}{n+1})$(1-$\frac{2}{n+1})$+⋯+($\frac{1}{{n+1}!})$ $2^n$(1-$\frac{1}{n+1})$(1-$\frac{2}{n+1})$...(1-$\frac{n}{n+1})$

For n≥2,$a_{n+1}$ >a_n Hence Monotone increasing.

$a_n ≤ 1+2+(2/2!)+(2/3!)+⋯

a_n ≤1+(2+(2/2^2) +(2/2^3) +⋯)

a_n ≤ 1+2(1+1/2^2 +1/2^3 +⋯)

By using sum to infinity, $a_n ≤1+2(2(1-1/2^n ))

$a_n ≤1+2(2-2^(1-n) )≤1+4-2^(2-n)≤5;as n→∞.

Hence convergent and bounded above by 5.

Tosh
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