Prove that the sequence $(s_n)$ is bounded above by 3 and increasing

Question

Let $(s_n)$ be the sequence defined by $s_n=(1+\frac{1}{n})^n$. Use the binomial theorem to show that $(s_n)$ is an increasing sequence with $s_n<3$ for all n.

This is a question that I am working on. I’m having a hard time with this proof. I think I can prove this result using induction. As the problem states I am required to use the binomial theorem. I have seen people use the Bernoulli inequality but I have to use the binomial theorem. Below is the work that I have done so far.

$(x+y)^n=\sum_{k=0}^{n} {n\choose k} x^{n-k} y^k$.

Applying the binomial theorem to $s_n$:

$(1+\frac{1}{n})^n = \sum_{k=0}^{n} {n\choose k} 1^{n-k} (\frac{1}{n})^k = \sum_{k=0}^{n} {n\choose k} (\frac{1}{n})^k$

${n\choose k} (\frac{1}{n})^k = \frac{1}{k!} \frac{n!}{n^k(n-k)!} = \frac{1}{k!} \frac{n(n-1)(n-2)…2*1}{n^k(n-k)(n-(k+1))…2*1} = \frac{1}{k!} \frac{(n-1)(n-2)…(n-(k-1))}{n^{k-1}}$

${n\choose k} (\frac{1}{n})^k = \frac{1}{k!} (1-\frac{1}{n})(1-\frac{2}{n})…(1-\frac{k-1}{n})$

Induction:

$\textbf{Base Case:}$ $n=1$

$s_1=\sum_{k=0}^{1} {1\choose k} = 1+1=2$

$s_2= \sum_{k=0}^{2} {2\choose k} (\frac{1}{2})^k= {2\choose 0} (\frac{1}{2})^0 + {2\choose 1} (\frac{1}{2})^1 + {2\choose 2} (\frac{1}{2})^2 = 1+1+\frac{1}{4}=\frac{9}{4}$

$s_1<3$, $s_2<3$, $s_1<s_2$

$\textbf{Induction Step:}$ If true for $s_k$, then true for $s_{k+1}$.

This is the part where I’m stuck in the proof. I want to use:

${n\choose k} (\frac{1}{n})^k = \frac{1}{k!} (1-\frac{1}{n})(1-\frac{2}{n})…(1-\frac{k-1}{n})$

but I’m confused because when k=0, k-1=-1 and when k=1, k-1=0. Can I use this when I is equal to these values? If not can you explain why because the derivation of this formula makes it seem like I can choose these values. Am I doing the proof incorrectly? Can you help me finish this proof?

I have shown all the work I have done and believe this question is specific enough to not get closed. If you need any clarification on anything that I am asking please feel free to ask. Thank you for your help!

Answer 1

As you correctly wrote $$ \left(1+\frac{1}{n}\right)^n =\sum_{k=0}^{n} {n\choose k} \frac{1}{n^k} =\sum_{k=0}^{n} a_{k,n} $$ where $$ a_{k,n}=\frac1{k!} \left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)...\left(1-\frac{k-1}{n}\right). $$ It is clear that each term in the product above is increasing in $n$, hence $a_{k,n}$ is increasing in $n$ for each fixed $k$. Consequently, $$ s_{n+1}=\left(1+\frac{1}{n+1}\right)^{n+1} =\sum_{k=0}^{n} a_{k,n+1}=\sum_{k=0}^{n+1} a_{k,n+1} =a_{n+1,n+1}+\sum_{k=0}^{n} a_{k,n+1}>\sum_{k=0}^{n} a_{k,n+1}> \sum_{k=0}^{n} a_{k,n}=\left(1+\frac{1}{n}\right)^n=s_n . $$

Finally, $a_{k,n}\le \frac 1{k!}$ so $$ s_n\le 1+\frac1{1!}+\frac1{2!}+\frac1{3!}+\dots \le 1+1+\frac12+\frac1{2^2}+\frac1{2^3}+\dots =2+\left[\sum_{k=1}^\infty \frac1{2^k}\right]=3 $$ using the fact that $k!=1\cdot2\cdot3\cdot\dots\cdot k\ge 2^{k-1}$.