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limit of $$\lim_{n\to \infty}\sqrt{n}\left(\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^{n}\right)=?$$

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What have you tried, wildcow? Have you expanded the middle out and gotten everything with a common denominator, for example? That is the first thing I would try, at least. –  Aaron Feb 2 '13 at 4:16
    
Write it in the form:0/0, and then use L'Hôpital. Notice that $\lim_{n\to \infty} (1+1/(n+1))^n=e$ still. –  awllower Feb 2 '13 at 4:24

2 Answers 2

$$\begin{align*}0\leq\sqrt{n}\left[\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^{n}\right]&\leq\sqrt{n}\left[\left(1+\frac{1}{n}\right)^{n+1}-\left(1+\frac{1}{n}\right)^{n}\right]\\&\;\;\;\;\;=\frac{1}{\sqrt{n}}\left(1+\frac{1}{n}\right)^{n}\\&\quad\;\leq \frac{e}{\sqrt{n}}\end{align*}$$

Hence:

$$0\leq \lim_{n\to\infty}\sqrt{n}\left[\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^{n}\right]\leq \lim_{n\to\infty}\frac{e}{\sqrt{n}}=0$$

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Nice (+1). Actually, your way helped me in finding an elementary proof for another problem. –  Chris's sis Feb 2 '13 at 12:12

$$\lim_{n\to \infty}\sqrt{n}\left(\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^{n}\right)=\lim_{n\to \infty}\sqrt{n}\left(1+\frac{1}{n}\right)^{n}\left(\frac{\left(1+\frac{1}{n+1}\right)^{n+1}}{\left(1+\frac{1}{n}\right)^{n}}-1\right)=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}\lim_{n\to\infty}\sqrt{n}\left(\frac{\left(1+\frac{1}{n+1}\right)^{n+1}}{\left(1+\frac{1}{n}\right)^{n}}-1\right)=e\lim_{n\to\infty}\left(\frac{\sqrt n}{n+1}\right)=0$$

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