# Bounded Sequences (II)

I came across the following problems during the course of my studying of real analysis:

Show that the sequence $(a_n)$ defined by $a_n = \left(1+ \frac{1}{n} \right)^{n}$ is bounded above by $3$.

I think we can use the binomial theorem. So $$a_n = \left(1+ \frac{1}{n} \right)^{n} = \sum_{k=0}^{n} \binom{n}{k} \left(\frac{1}{n} \right)^{k}$$

$$= 1+ \sum_{k=1}^{n} \binom{n}{k} \left(\frac{1}{n} \right)^{k}$$

From here, how would I deduce that this is $\leq 3$?

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Like your last question this is also a classic one. – Américo Tavares Jun 22 '11 at 23:32
It would have been nice to see somewhat more computation. You mentioned an idea that you thought might be helpful. Indeed it is. But your questions are posted at such a fast rate that I think you may not be giving yourself enough time to seriously tackle each problem before seeking help. – André Nicolas Jun 22 '11 at 23:44

Just expand out the binomial coefficient as $$\frac{(n)(n-1)\cdots (n-k+1)}{k!}$$

Then you can conclude quickly that the sum is no greater than $$1+\frac{1}{1!}+\frac{1}{2!}+ \cdots$$

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So the sum is $$1+\sum_{k=1}^{n} \frac{(n)(n-1) \cdots (n-k+1)}{k! n^{k}}$$. – Damien Jun 23 '11 at 0:03
@Damien: Yes it is. But now note that in any term of the sum, the top is $\le n^k$, so the "typical" term in the sum is $\le \frac{1}{k!}$. That's how I get my second displayed line. And then one still has to show that that second displayed sum is less than $3$. That's easy. Third term is $1/2$, fourth is $<(1/2)^2$, fourth is $<(1/2)^3$, and so on. Maybe you will recognize we have in fact shown that your exponential is $<e$. – André Nicolas Jun 23 '11 at 0:14
Or one can show that $$1+ \frac{1}{1!} + \frac{1}{2!} + \cdots + \frac{1}{n!} \leq 2$$ by noting that $n! \geq 2^{n-1}$ for all $n$. – Damien Jun 23 '11 at 0:19
@Damien: That's exactly what my informal "Third term is $1/2$, fourth is $<(1/2)^2$, fourth is $<(1/2)^3$ and so on says". – André Nicolas Jun 23 '11 at 0:27