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I'm kind of confused, because I'm not sure how this expression $$ \sum_{k=0}^n \frac{n!}{k!(n-k)!n^k} $$ simplifies to $$ \sum_{k=0}^n \frac{1}{k!} $$

Could anybody help me out?

Michael Hardy
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Matt24
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  • What you've written doesn't make much sense as the "simplified expression" has $n$ as the index so it goes to $\infty$. – Klint Qinami Aug 20 '16 at 18:05
  • It does not. Moreover (though it is not the crucial fact), the second sum is $+\infty$, because $k$ does not depend on $n$. –  Aug 20 '16 at 18:05
  • "k" is the index in both expressions, my bad. Just edited it. – Matt24 Aug 20 '16 at 18:11
  • In the first expression the sum goes up to $n$. In the limit $n\to\infty$ it converges to the second expression. Also each term ${n \choose k} \frac{1}{n^k}$ converges to $\frac{1}{k!}$. – sometempname Aug 20 '16 at 18:15
  • @sometempname Hold on, how is the limit of the first expression 1/k! ? – Matt24 Aug 20 '16 at 18:25
  • In the second expression the sum goes to $\infty$. For each fixed $k$ you have ${n \choose k} \frac{1}{n^k} \to \frac{1}{k!}$ as $n\to\infty$, and the finite sum also converges to the infinite sum. There is no identity\simplification for any finite $n$ (except $k=0,1$). – sometempname Aug 20 '16 at 18:30
  • These expressions are not equal. Even with $n=2$, we get $\frac94$ and $\frac52$. – robjohn Aug 20 '16 at 18:37
  • @sometempname This is at best a heuristic argument, though. When $n\to\infty$, you change both the summands and the number of terms of the first sum, while the number of terms of the second one changes as well -- this is a good intuition, maybe, but barely more. – Clement C. Aug 20 '16 at 19:32
  • Ah, true. I had written down that they were equal in class, but surely then I must've forgotten to add that their limit was the same.

    Thanks for the help everyone!

     – Matt24 Aug 20 '16 at 19:33
  • I didn't give any argument, just stated the facts. – sometempname Aug 20 '16 at 19:36
  • @sometempname "In the limit $n\to\infty$ it converges to the second expression." No, it doesn't, since the second expression also has a $n$ in it. – Clement C. Aug 20 '16 at 19:38
  • Like I wrote above, the second sum should go to $\infty$ (the original post was different from the current one). – sometempname Aug 20 '16 at 19:44

2 Answers2

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As Clement C. already noted, the two expressions are not equivalent. Here, we show that, in fact, $\left(1+\frac1n\right)^n\le \sum_{k=0}^n \frac{1}{k!}$ for all $n\ge 1$. Moreover, we show that as $n\to \infty$, both terms of interest approach the same value. To that end, we proceed.

First, note that using the binomial theorem, we can write

$$\begin{align} \left(1+\frac1n\right)^n&=\sum_{k=0}^n \binom{n}{k}\frac{1}{n^k}\\\\ &=\sum_{k=0}^n \frac{1}{k!}\frac{n!}{n^k(n-k)!}\\\\ &=\sum_{k=0}^n \frac{1}{k!}\frac{n(n-1)(n-2)\cdots (n-k+1)}{n^k}\\\\ &=\sum_{k=0}^n \frac{1}{k!}\left(\frac nn\right)\left(\frac{n-1}{n}\right)\left(\frac{n-2}{n}\right)\cdots \left(\frac{n-k+1}{n}\right)\\\\ &=\sum_{k=0}^n \frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{k-1}{n}\right)\tag 1 \end{align}$$

Second, since all of the parenthetical terms in $(1)$ are less than $1$, we arrive at the first coveted relationship

$$\bbox[5px,border:2px solid #C0A000]{\left(1+\frac1n\right)^n \le \sum_{k=0}^n\frac{1}{k!}} \tag 2$$

for all $n\ge 1$.

Next, note that since all of the terms in the binomial expansion are positive, we have for any integer $0\le p<n$

$$\left(1+\frac1n\right)^n \ge \sum_{k=0}^p \binom{n}{k}\frac{1}{n^k} \tag 3$$

Putting $(2)$ and $(3)$ together yields

$$\sum_{k=0}^p \binom{n}{k}\frac{1}{n^k} \le \left(1+\frac1n\right)^n \le \sum_{k=0}^n\frac{1}{k!} \tag 4$$

Since $\left(1+\frac1n\right)^n$ is monotonically increasing (SEE THIS ANSWER) and bounded above by the convergent series $\sum_{k=0}^\infty \frac{1}{k!}$, its limit exists. Similarly, the sum on the left-hand side of $(4)$ is monotonically increasing in $n$ and bounded above. Therefore, we may take the limit of $(4)$. Proceeding, we find

$$\sum_{k=0}^p \frac{1}{k!} \le \lim_{n\to \infty}\left(1+\frac1n\right)^n \le \sum_{k=0}^\infty\frac{1}{k!} \tag 5$$

for all $p$. Since the inequality on the left-hand side of $(5)$ is true fo all $p$, we can let $p\to \infty$ to reveal the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\left(1+\frac1n\right)^n=\sum_{k=0}^\infty \frac{1}{k!}}$$

Mark Viola
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tl;dr: the two are not equal, but have the same limit when $n$ goes to $\infty$.

First: You cannot "simplify the first to get the second." The two expressions are not equal.

Compute the first few terms to check that. I am lazy, so asked a software program to do it for me. The ratio of their first 10 terms (obtained with Mathematica):

 N[Table[(Sum[1/(k!), {k, 0, n}]/Sum[n!/(k! (n - k)! n^k), {k, 0, n}]), {n, 1, 10}] ]

{1., 1.11111, 1.125, 1.10933, 1.09177, 1.0779, 1.06745, 1.05943, 1.05312, 1.04802}

A plot (ditto)

DiscretePlot[{Sum[1/(k!), {k, 0, n}], Sum[n!/(k! (n - k)! n^k), {k, 0, n}]}, {n, 1, 100}]

enter image description here

However, the two expressions do have the same limit when $n\to\infty$, namely $e$.

For the first, it is trivial based on the series definition of $e$: $$ \sum_{k=0}^n \frac{1}{k!} \xrightarrow[n\to\infty]{} \sum_{k=0}^\infty \frac{1}{k!} = e $$

For the second, observe that by the Binomial theorem $$ \sum_{k=0}^n \frac{n!}{k!(n-k)!} \frac{1}{n^k} = \sum_{k=0}^n \binom{n}{k}\frac{1}{n^k} = \left(1+\frac{1}{n}\right)^n \xrightarrow[n\to\infty]{} e $$ where the last step is a standard limit (or, for some, this is the definition of $e$).

Clement C.
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