Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given $$ e=\sum\limits_{k=0}^\infty\frac{1}{k!} $$ How can I prove $$ e^n=\sum\limits_{k=0}^\infty\frac{n^k}{k!} $$

Can anyone please demostrate the $n=2$ case? Thanks!

share|improve this question

4 Answers 4

up vote 11 down vote accepted

$$e^2=\sum_{k=0}^\infty\sum_{m=0}^\infty\frac{1}{k!}\frac{1}{m!}=\sum_{k=0}^\infty\sum_{m=0}^\infty\frac{(m+k)!}{k!m!}\frac{1}{(m+k)!}$$

Now, denote $l:=m+k$ and group the terms by $l$

$$e^2=\sum_{l=0}^\infty\sum_{k=0}^l\frac{l!}{k!(l-k)!}\frac{1}{l!}=\sum_{l=0}^\infty\left[\sum_{k=0}^l\binom{l}{k}\right]\frac{1}{l!}=\sum_{l=0}^\infty\left[2^l\right]\frac{1}{l!}$$

share|improve this answer
5  
It can be useful to remark that this is true thanks to the absolute convergence of the series that defines $e$. More generally, this follows from a convergence theorem for Cauchy products of series. –  Siminore May 11 '12 at 14:46
    
The multinomial generalization to all natural $n$ should be clear from this answer. –  anon May 11 '12 at 14:55

We can work by induction. The base case $n=1$ is trivially true. Suppose it is true for $n$, then

$$e^n =\sum_{i=0}^\infty \frac{n^i}{i!}$$

$$e^n e =\sum_{k=0}^\infty \frac{1}{k!} \sum_{i=0}^\infty \frac{n^i}{i!}$$

$${e^{n + 1}} = \sum\limits_{k = 0}^\infty {\sum\limits_{i = 0}^\infty {\frac{1}{{k!i!}}} } {n^i} = \sum\limits_{k = 0}^\infty {\sum\limits_{i = 0}^\infty {\frac{{\left( {k + i} \right)!}}{{k!i!}}} } \frac{{{n^i}}}{{\left( {k + i} \right)!}}$$

We procede with $m=k+i$ to get $${e^{n + 1}} = \sum\limits_{k = 0}^\infty {\sum\limits_{i = 0}^\infty {\frac{1}{{k!i!}}} } {n^i} = \sum\limits_{m = 0}^\infty {\sum\limits_{i = 0}^m {\frac{{m!}}{{\left( {m - i} \right)!i!}}} } \frac{{{n^i}}}{{m!}}$$ $${e^{n + 1}} =\sum\limits_{m = 0}^\infty {\left( {\sum\limits_{i = 0}^m {{m\choose i}{n^i}} } \right)} \frac{1}{{m!}}$$

$${e^{n + 1}} = \sum\limits_{m = 0}^\infty {\frac{{{{\left( {n + 1} \right)}^m}}}{{m!}}} $$

Note that we could have left $\infty$ as the upper limit instead of $m$, since the binomial theorem is a special case of the general binomial theorem. Also note that the change in the index of summation follows the relation established by $k+n=r$. Since the hypothesis is true for $n=1$, and $k=n \rightarrow k=n+1$, the formula holds for every $n$ a natural number.

share|improve this answer

The easiest way, i think, is to use Tayor's expansion formula for $e^x$, and then replace $x$ by $n$ and get that:

$$e^n=\sum\limits_{k=0}^\infty\frac{n^k}{k!}$$

The proof is complete.

share|improve this answer

Use $\displaystyle\frac{d}{dn}e^n=e^n$ and derive the sum, which is equivalent to the sum itself, so it's equivalent to $e^n$.

share|improve this answer
    
It seems $n$ is an integer here. –  Pedro Tamaroff Jul 12 '13 at 6:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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