$a_n=(1+\frac{1}{n})^n$ and $n\geq 1$ I check first $n\to \infty$ then $\lim_{n\to \infty}=(1+\frac{1}{\infty})^\infty$ $=(1+0)^\infty=1^\infty$ I don't find the next step.Help me
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It is well known that this sequence converges to $e$. This is a perfect example of why you can not just evaluate the limit at different points of the expression separately. $\lim\limits_{x\to\infty}\left(f(x)^{g(x)}\right)$ is in general not equal to $\lim\limits_{x\to\infty}\left((\lim\limits_{y\to\infty}f(y))^{g(x)}\right)$ – JMoravitz Oct 14 '17 at 21:45
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Which series are you speaking of? As to the $(a_n)$ sequence, it is a standard result that its limit is Euler's number $\mathrm e$. – Bernard Oct 14 '17 at 21:45
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If all you wish to do is prove convergence, you can prove the following two things: It is a monotonic increasing sequence, it is bounded above (for example, by $3$). These two together will by the dominated convergence theorem imply the convergence of the sequence. To prove that it specifically converges to $e$, this will depend on which specific definition of $e$ you are using and will utilize the properties of $e$. – JMoravitz Oct 14 '17 at 21:47
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I know the result equals $e$ but i don't show.How can I show? – burak kılıç Oct 14 '17 at 21:50