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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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  • $\begingroup$ Could you define $e'$? $\endgroup$
    – Jacob
    Commented Nov 14, 2011 at 16:47
  • $\begingroup$ I believe it is defined int his homework as just $e$, I will change the OP. $\endgroup$ Commented Nov 14, 2011 at 16:54
  • $\begingroup$ use $\lim_{n \to \infty} (1+\frac{1}{n})^{n}=e$ $\endgroup$
    – Pedja
    Commented Nov 14, 2011 at 16:54
  • $\begingroup$ @pedja: I think you need the stronger $\lim_{n \rightarrow \infty} (1 + \frac{x}{n})^n = e^x$. (Applied with $x = -1$.) $\endgroup$ Commented Nov 14, 2011 at 16:59
  • $\begingroup$ Assuming you are to establish the limit, L' Hospital rule may be a good choice. $\endgroup$
    – Tapu
    Commented Nov 14, 2011 at 17:04

1 Answer 1

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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.

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    $\begingroup$ Nice solution, uses almost nothing other than the definition of $e$. $\endgroup$ Commented Nov 14, 2011 at 17:17

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