3
$\begingroup$

I wish to show that $\lim_{n\to\infty}\left(1+\frac{\lambda}{n}\right)^n=e^\lambda$ as part of the proof of the Poisson limit theorem. My first avenue of attack was to take the log of both sides and then use L'Hopital's rule, but the result was of the form $0\cdot \infty$, so that didn't work.

I also noticed that the proof would follow from the proof that $\left(1+\frac{1}{n}\right)^\lambda = \left(1+\frac{\lambda}{n}\right)$ for large $n$, but I wasn't able to prove this either.

Any hints?

$\endgroup$

marked as duplicate by Guy Fsone, mechanodroid, Namaste calculus Nov 11 '17 at 0:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 7
    $\begingroup$ What is your definition of $e^\lambda$? $\endgroup$ – Argon Jan 21 '13 at 15:10
  • 4
    $\begingroup$ $\left(1+\frac\lambda n\right)^n = \big(\left(1+\frac\lambda n\right)^{n/\lambda}\big)^\lambda$, and $\lim_{x\to\infty}(1+\frac1x)^x = e$. $\endgroup$ – Berci Jan 21 '13 at 15:14
  • $\begingroup$ @Argon: $\lim_{n\to\infty}(1+\frac{1}{n})^{n\lambda}$ $\endgroup$ – Xodarap Jan 21 '13 at 15:16
  • $\begingroup$ The form $0 \cdot \infty$ is easy to convert to $0 / 0$ or $\infty / \infty$. $\endgroup$ – Hurkyl Jan 21 '13 at 16:16
7
$\begingroup$

If $$y=\left(1+\frac \lambda n\right)^n$$

so, $$\ln y=n\ln\left(1+\frac \lambda n\right)=\frac{\ln\left(1+\frac \lambda n\right)}{\frac1n}$$

so, $$\lim_{n\to\infty}\ln y=\lim_{n\to\infty}\frac{\ln\left(1+\frac \lambda n\right)}{\frac1n}\text{ which is in the form } \frac00 $$

Applying L'Hospital's rule, $$\lim_{n\to\infty}\ln y=\lim_{n\to\infty}\frac{-\frac \lambda {n^2}}{-\frac1{n^2}\left(1+\frac \lambda n\right)}=\lambda$$

So, $$\lim_{n\to\infty}\ln y=\lambda\iff \lim_{n\to\infty}=e^\lambda$$

$\endgroup$
  • 3
    $\begingroup$ You forgot the logarithm in the numerator on both step 2 and step 3, which leads to a wrong derivative of the numerator in step 4. $\endgroup$ – DoomMuffins Jan 22 '13 at 19:18
  • $\begingroup$ @DoomMuffins, thanks a lot. Could you please have a look into the edited version? $\endgroup$ – lab bhattacharjee Jan 23 '13 at 3:10
  • $\begingroup$ Well rectified. $\endgroup$ – DoomMuffins Jan 23 '13 at 22:10
  • $\begingroup$ @labbhattacharjee: cool proof by the way! $\endgroup$ – Alex Jan 24 '13 at 21:08
6
$\begingroup$

The simple and straightforward way is to use substitution $n=\lambda t$. Your limit becomes: $$\lim_{t\to\infty}\left(1+\frac{1}{t}\right)^{\lambda t}=\left(\lim_{t\to\infty}\left(1+\frac{1}{t}\right)^t\right)^\lambda=e^\lambda$$

$\endgroup$
5
$\begingroup$

The answer very much depends on what you know, but, if this includes the fact that log is a primitive of $x\mapsto1/x$ and the inverse of $x\mapsto\mathrm e^x$, you may start from $$ 1-x\leqslant\frac1{1+x}\leqslant 1. $$ Integrating this from $0$ to some $x\geqslant0$ yields $$ x-\tfrac12x^2\leqslant\log(1+x)\leqslant x, $$ which implies $$ \mathrm e^{x-x^2/2}\leqslant1+x\leqslant\mathrm e^x, $$ and, for every nonnegative $n$, $$ \mathrm e^{nx-nx^2/2}\leqslant(1+x)^n\leqslant\mathrm e^{nx}. $$ Using this for $x=\lambda/n$ yields the nonasymptotic bounds $$ \mathrm e^{\lambda-\lambda^2/(2n)}\leqslant(1+\lambda/n)^n\leqslant\mathrm e^{\lambda}. $$ Since $\lambda^2/(2n)\to0$, you are done.

$\endgroup$
3
$\begingroup$

$$e = \lim_{n \to \infty} \left(1+\frac 1 n\right)^n$$

$$e^\lambda = \lim_{n \to \infty} \left(1+\frac 1 n\right)^{n \lambda} =\\ \lim_{n \to \infty} \left( \left(1+\frac 1 n\right)^\lambda\right)^{n}=\\ \lim_{n \to \infty} \left(1+\frac{\lambda}{n}+O\left(\frac{1}{n^2}\right)\right)^{n} $$

$\endgroup$
2
$\begingroup$

We know, $$e^y=\sum_{0\le r<\infty}\frac{y^r}{r!}$$

The $r$th term $T_r$ of $\left(1+\frac\lambda n\right)^n$ is $\frac {n(n-1)(n-2)\cdots\{n-(r-1)\}}{1\cdot2\cdot3\cdots (r-1)r}\left(\frac\lambda n\right)^r$ for $r\ge 1, T_0=1$

So, $\lim_{n\to\infty}T_r$ of $\left(1+\frac\lambda n\right)^n$ is $$\lim_{n\to\infty}\frac {n(n-1)(n-2)\cdots(n-r+1)}{1\cdot2\cdot3\cdots (r-1)r}\left(\frac\lambda n\right)^r$$

$$=\frac{\lambda^r}{r!}\lim_{n\to\infty}\prod_{0\le s\le r-1}\left(1-\frac sn\right)=\frac{\lambda^r}{r!}$$

So, $$\lim_{n\to\infty}\left(1+\frac\lambda n\right)^n=1+\sum_{1\le r<\infty}\frac{\lambda^r}{r!}=\sum_{0\le r<\infty}\frac{\lambda^r}{r!}=e^\lambda$$

$\endgroup$
  • $\begingroup$ Summations in series are never over $r\leqslant\infty$ but over $r\lt\infty$. More importantly, the sum of the limits of the terms of a series is not always the limit of the sum of this series. $\endgroup$ – Did Jan 22 '13 at 19:03
  • $\begingroup$ @Did, could you please elaborate the last statement? $\endgroup$ – lab bhattacharjee Jan 22 '13 at 19:07
  • $\begingroup$ Sure: $x_n(k)\to z_n$ when $k\to\infty$, for each $n$, does not imply that $\sum\limits_{n=1}^{+\infty}x_n(k)\to\sum\limits_{n=1}^{+\infty}z_n$ when $k\to\infty$. $\endgroup$ – Did Jan 22 '13 at 19:51
  • $\begingroup$ @Did, if we are allowed to use $e^y=\sum_{0\le r<\infty}\frac{y^r}{r!},$ can't we write $\lim_{n\to\infty}\left(1+\frac\lambda n\right)^n=\sum_{0\le r<\infty}\frac{\lambda^r}{r!}=e^\lambda?$ $\endgroup$ – lab bhattacharjee Jan 23 '13 at 6:14
  • $\begingroup$ The question is why could we? Unless you address this question, your method is incomplete. $\endgroup$ – Did Jan 23 '13 at 6:36

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