I have to find the limit of a sequence $(1+1/n)^{(1+n)}$. Any help is most welcome.
Limit of a sequence $(1+1/n)^{(n+1)}$
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8$(1 + \frac{1}{n})^{n + 1} = (1 + \frac{1}{n})(1 + \frac{1}{n})^n$ – 2012-06-27
2 Answers
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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0Thanks Siminore! Yes I know the famous limit and now this one becomes very simple to compute. – 2012-06-27
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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0Don't worry, I appreciate irony! :-) – 2012-06-27