Limit of a sequence $(1+1/n)^{(n+1)}$

I have to find the limit of a sequence $(1+1/n)^{(1+n)}$. Any help is most welcome.

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I assume you want the limit as $n \to \infty$. Are you familiar with $$\lim_{n \to \infty} \left( 1+ \dfrac1n \right)^n = e?$$ – user17762 Jun 27 '12 at 7:36
It is $e$.Power $(n+1)$ doesn't change anything in the limit given by Marvis. – Aang Jun 27 '12 at 7:38
$(1 + \frac{1}{n})^{n + 1} = (1 + \frac{1}{n})(1 + \frac{1}{n})^n$ – William Jun 27 '12 at 7:38

This is not a complete answer, but a big hint: do you agree that $$\left( 1 + \frac{1}{n} \right)^{n+1} = \left( 1 + \frac{1}{n} \right)^n \left( 1 + \frac{1}{n} \right) \ ?$$ As $n \to +\infty$, can you read a very famous limit on the right-hand side? The other term won't do you any harm, since it converges to ...

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Thanks Siminore! Yes I know the famous limit and now this one becomes very simple to compute. – rajendra bakre Jun 27 '12 at 8:59

Use $$\left( 1 + \frac{1}{n} \right)^{n+1} =\exp\left((n+1)\log(1+\frac{1}n)\right)$$ and look at the limit for the exponent: $$\lim_{n\to \infty} \left((n+1)\log(1+\frac{1}n)\right)=\lim_{n\to \infty} \frac{\log(1+\frac{1}n)}{\frac{1}{n+1}}=\frac{0}{0}$$ so we can try L'Hospitâl and get $$\lim_{n\to \infty} \frac{-\frac{1}{n^2}\frac{1}{1+1/n}}{-\frac{1}{(n+1)^2}}=\lim_{n\to \infty} \frac{n+1}{n}=1.$$ Therefore your limit is $e^1$.

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As I always remark: you need to apply De l'Hospital to the formal extension of the given sequence to the set of real numbers. It may seem a pedantic issue, but it isn't. – Siminore Jun 27 '12 at 8:24
thanks draks! I got the point – rajendra bakre Jun 27 '12 at 8:57
@Siminore: Can you elaborate a bit more? I didn't understand your point... Thanks – rubik Jun 27 '12 at 9:14
@Siminore do you mean, one should mention that $n$ is element of the Extended real number line $\Bbb{R}\cup \pm \infty$? – draks ... Jun 27 '12 at 9:30
It is a trivial fact, and I know that many users don't like it. I just mean that you cannot apply De l'Hospital to a sequence $a_n/b_n$. You can if you know that $a_n=f(n)$ and $b_n=g(n)$, where $f$ and $g$ satisfy De l'Hospital's theorem. In this case it is clear. But I have seen things, as a teacher, that you cannot imagine :-) I've seen differentiation of $(-1)^n$ with respect to $n$, for instance. – Siminore Jun 27 '12 at 9:31