This is a homework question and I am not really sure where to go with it. I have a lot of trouble with sequences and series, can I get a tip or push in the right direction?
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Could you define $e'$? – Jacob Nov 14 '11 at 16:47
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I believe it is defined int his homework as just $e$, I will change the OP. – Matt Nashra Nov 14 '11 at 16:54
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use $\lim_{n \to \infty} (1+\frac{1}{n})^{n}=e$ – Pedja Nov 14 '11 at 16:54
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@pedja: I think you need the stronger $\lim_{n \rightarrow \infty} (1 + \frac{x}{n})^n = e^x$. (Applied with $x = -1$.) – Michael Joyce Nov 14 '11 at 16:59
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Assuming you are to establish the limit, L' Hospital rule may be a good choice. – Tapu Nov 14 '11 at 17:04
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@Robjohn: I wonder if your comment is confusing in the current state of the post :) (since the accepted answer has been changed) – SBF Mar 07 '12 at 11:29
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@Ilya: deleted more like. I guess that my comment should have been to that answer. We should leapfrog deletes :-) – robjohn Mar 07 '12 at 12:12
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You have: $$ x_n:=\left(1-\frac1n\right)^{-n} = \left(\frac{n-1}n\right)^{-n} = \left(\frac{n}{n-1}\right)^{n} $$ $$ = \left(1+\frac{1}{n-1}\right)^{n} = \left(1+\frac{1}{n-1}\right)^{n-1}\cdot \left(1+\frac{1}{n-1}\right) = a_n\cdot b_n. $$ Since $a_n\to \mathrm e$ and $b_n\to 1$ you obtain what you need.
SBF
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16Nice solution, uses almost nothing other than the definition of $e$. – André Nicolas Nov 14 '11 at 17:17