# How do you determine whether the sequence a_n=2^n-n^2 converges, if so how do you find the limit?

Jan 7, 2017

${\lim}_{n \to \infty} {2}^{n} - {n}^{2} = \infty$

#### Explanation:

Using the binomial expansion we have that:

${2}^{n} = {\left(1 + 1\right)}^{n} = {\sum}_{k = 0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) {1}^{n} \cdot {1}^{n} = {\sum}_{k = 0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right)$

If we consider $n > 3$ and isolate the first three terms:

${2}^{n} = 1 + n + n \left(n - 1\right) + {\sum}_{k = 3}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) = 1 + {n}^{2} + {\sum}_{k = 3}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right)$

Or:

${2}^{n} - {n}^{2} = 1 + {\sum}_{k = 3}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right)$

As:

$\left(\begin{matrix}n \\ k\end{matrix}\right) \ge 1$

we have:

${\sum}_{k = 3}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) \ge {\sum}_{k = 3}^{n} 1 = n + 1 - 3 = n - 2$

So:

${2}^{n} - {n}^{2} \ge n - 2$

and then:

${\lim}_{n \to \infty} {2}^{n} - {n}^{2} = \infty$