TITLE: How can I prove this sequence converges to 1?
QUESTION [5 upvotes]: Suppose $0<a<1$ and define $a_n=(1+a^n)^n$, show that $a_n \to 1$ by using binomial expansion on each $a_n$ and compare to a geometric sum.
I know how to compute binomial on $a_n$ but I am really confused by "using binomial expansion on each $a_n$ and compare to a geometric sum".

REPLY [6 votes]: I guess they've meant "to compare to a geometric sum", here's what you can do:
\begin{align} (1+a^n)^n &= \sum_{k=0}^n \binom{n}{k} a^{nk} = \sum_{k=0}^n \frac{n(n-1)\dots(n-k+1)}{k!} a^{nk} \le \\
&\le \sum_{k=0}^n \frac{n^k}{k!}a^{nk} \le \sum_{k=0}^n n^ka^{nk} \le \sum_{k=0}^\infty (na^n)^k = \frac{1}{1-na^n}\end{align}
For $0<a<1$ we have $$\lim_{n\rightarrow\infty} na^n = 0$$so $$ \lim_{n\rightarrow\infty} \frac{1}{1-na^n} = 1 $$
and using the squeezing $$ 1\le a_n \le \frac{1}{1-na^n}$$
we get $$ \lim_{n\rightarrow\infty} a_n = 1$$