Show $\lim_{n\to \infty} 2^n-(2+\frac{1}{n})^n = -\infty$ Show: 
$$\lim_{n\to \infty} a_n=-\infty, \qquad a_n=2^n-\left(2+\frac{1}{n}\right)^n$$ 
attempt.
$$2^n-\left(2+\frac{1}{n}\right)^n=2^n-2^n\left(1+\frac{1}{2n}\right)^n=2^n\left(1-\left(1+\frac{1}{2n}\right)^n\right)$$LEMMA: $(1+\frac{1}{2n})^n$ converges to some $\alpha >1$.  
My struggle is with proving it.
 A: 
We don't have to fine the limit of $\left(1+\frac{1}{2n}\right)^n$.  Rather, we only need to show that there exists a number $C>1$, such that for every $n$, $\left(1+\frac{1}{2n}\right)^n>C$.  To that end, we proceed.


Using Bernoulli's Inequality, we have 
$$\left(2+\frac1n\right)^n=2^n\left(1+\frac1{2n}\right)^n\ge \left(\frac32\right)\,2^n$$
Hence, 
$$2^n-\left(2+\frac1n\right)^n\le -2^{n-1}$$
whence letting $n\to\infty$ yields the coveted limit.
A: HINT
Recall that
$$\left(1+\frac{a}{n}\right)^n\to e^a$$
indeed
$$\left(1+\frac{a}{n}\right)^n=\left[\left(1+\frac{a}{n}\right)^\frac n a\right]^a$$
A: Hint: $$\left(2+\frac1n\right)^n=\left(2+2\cdot\frac{1/2}{n}\right)^n=2^n\left(1+\frac{1/2}{n}\right)^n$$
Now factor out the $2^n$ and determine what remains.
A: Hint: by the Binomial Theorem, for $n \ge 1$ we have
$$ \left( 2 + \frac{1}{n} \right)^n = \sum_{i=0}^n \binom{n}{i} 2^{n-i} \left( \frac{1}{n} \right)^i \ge 2^n + 2^{n-1}. $$
(The inequality is gotten from observing the sum is at least equal to the sum of the terms from $i=0$ and $i=1$.)
