18
$\begingroup$

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?

$\endgroup$
7
  • $\begingroup$ Could you define $e'$? $\endgroup$
    – Jacob
    Nov 14, 2011 at 16:47
  • $\begingroup$ I believe it is defined int his homework as just $e$, I will change the OP. $\endgroup$ Nov 14, 2011 at 16:54
  • $\begingroup$ use $\lim_{n \to \infty} (1+\frac{1}{n})^{n}=e$ $\endgroup$
    – Peđa
    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 $x = -1$.) $\endgroup$ Nov 14, 2011 at 16:59
  • $\begingroup$ Assuming you are to establish the limit, L' Hospital rule may be a good choice. $\endgroup$
    – Tapu
    Nov 14, 2011 at 17:04

1 Answer 1

62
$\begingroup$

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.

$\endgroup$
1
  • 16
    $\begingroup$ Nice solution, uses almost nothing other than the definition of $e$. $\endgroup$ Nov 14, 2011 at 17:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.