I've looked up several related questions, but they do not answer what I am looking for. Please give link if this is a duplicate.

What I eventually want to know is why


converges to a finite value.

I mean, we have to know that the limit converges before we can call it a constant, right?

I know that $f(n)=\left(1+\dfrac1n\right)^n$ is strictly increasing, but how do we know it is bounded above?

Also, I tried using the binomial theorem to write

$$\lim_{n\to\infty}\left(1+\frac1n\right)^n=\lim_{n\to\infty}\left[\sum_{k=0}^n\frac{n!}{n^k k! (n-k)!}\right],\tag{1}$$

but I am unsure how to prove that the sum converges for $n\to\infty,$ because the general term depends on both $n$ and $k.$

Here is what I think will work.

Let $a_{n,k}=\dfrac{n!}{n^k k! (n-k)!}.$ Then $a_{n+1,k+1}=\dfrac{(n+1)!}{(n+1)^{k+1} (k+1)! (n-k)!}.$

Using the Ratio Test, we get

$$\lim_{n\to\infty\atop k\to\infty}\left|\frac{a_{n+1,k+1}}{a_{n,k}}\right|=\lim_{n\to\infty\atop k\to\infty}\left[\frac{1}{k+1}\left(\frac{n}{n+1}\right)^k\right].$$

where the limit on the RHS is less than 1.

That may take care of the case where $n\in\mathbb{N},$ so I suppose that (1) can be generalized using "Newton's generalised binomial theorem."

  • $\begingroup$ @martinsleziak, and how would I begin that? I don't see how your comment helps. $\endgroup$ – John Molokach Jun 22 '15 at 14:10
  • $\begingroup$ My comment was posted in order to provide you (and other readers of the question) with links to some very similar questions that have been asked before on the site. I was able to find two of them, maybe there is much more than that. The links are now displayed also in the sidebar on the right among the linked questions. $\endgroup$ – Martin Sleziak Jun 22 '15 at 14:14
  • $\begingroup$ Sorry failed to see your comment was a link. $\endgroup$ – John Molokach Jun 22 '15 at 14:15
  • $\begingroup$ Sorry failed to see your comment was a link. Of the two links, the main answers are a repeat of the one posted below. The one exception being the use of the Binomial Theorem in comparison to an expression that decomposes into partial fractions, which is not quite clear to me. $\endgroup$ – John Molokach Jun 22 '15 at 14:20

$$\begin{align}\left(1+\frac 1n\right)^n&=\sum_{k=0}^{n}\binom{n}{k}\left(\frac 1n\right)^k\\&=\sum_{k=0}^{n}\frac{n(n-1)\cdots(n-k+1)}{k!}\left(\frac 1n\right)^k\\&=\sum_{k=0}^{n}\frac{1}{k!}\left(1-\frac 1n\right)\left(1-\frac 2n\right)\cdots\left(1-\frac{k-1}{n}\right)\\&\lt \sum_{k=0}^{n}\frac{1}{k!}\\&=1+\frac{1}{1}+\frac{1}{2}+\frac{1}{3\cdot 2}+\frac{1}{4\cdot 3\cdot 2}+\cdots+\frac{1}{n\cdot(n-1)\cdots 2}\\&\lt 1+\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{n-1}}\\&=3-\frac{1}{2^{n-1}}\\&\lt 3\end{align}$$

  • 1
    $\begingroup$ Line 4 to line 5 didn't make sense to me at first glance, so I suggested an edit. Very nice! I want to let this simmer a bit to see how many different ways people come up with for the upper bound. Also, does the method I proposed in the question work? $\endgroup$ – John Molokach Jun 22 '15 at 12:50

First you should know that: $$ (1) \ln(1+x)=x-\frac{x^2}{2}+O(x^3)$$ Also consider that ln is an injective function meaning if $\ln(A)=1$ then $A=e$
suppose $A=(1+\frac{1}{n})^n$ So: $$ \lim_{n\to\infty} \ln(A) = \lim_{n\to\infty} \ln((1+\frac{1}{n})^n = \lim_{n\to\infty} n\ln(1+\frac{1}{n}) $$ Applying (1): $$ \lim_{n\to\infty} n\ln(1+\frac{1}{n}) = \lim_{n\to\infty} n(\frac{1}{n}-\frac{1}{2n^2}+O(\frac{1}{n^3}))$$ $$\lim_{n\to\infty} n(\frac{1}{n}-\frac{1}{2n^2}+O(\frac{1}{n^3})) = \lim_{n\to\infty} (1-\frac{1}{2n}+nO(\frac{1}{n^3})) = 1 $$ Since $$ \lim_{n\to\infty} \ln(A) = 1 $$ Than $$ \lim_{n\to\infty} A = e $$ So e is the least upper bound.


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