Limit of a Sequence 001

Consider the expression:

$N = \sqrt{4n^2 + 5n + 1}$

$N = 2n\sqrt{1 + \frac{5}{4n} + \frac{1}{4n^2}}$

If $n$ is a very large integer, then the terms $\frac{5}{4n}$ and $\frac{1}{4n^2}$ are very small. This allows us to approximate $N$ as follows:

$N = 2n\left(1 + \frac{5}{4n} + \frac{1}{4n^2}\right)^{1/2}$

$N \approx 2n\left(1 + \frac{1}{2}\left(\frac{5}{4n} + \frac{1}{4n^2}\right)\right)$ $\implies N \approx 2n + \frac{5}{4} + \frac{1}{4n}$

The above is a first-degree binomial approximation of the expression. Now, the cosine can also be approximated as such:

$\cos(\pi N) = \cos\left(2\pi n + \frac{5 \pi}{4} + \frac{\pi}{4n}\right)$

Since $n$ is large, then $\frac{\pi}{4n}$ can be neglected. Then:

$\cos(\pi N) = \cos\left(2\pi n + \frac{5 \pi}{4}\right)$

Since $n$ is an integer:

$\cos(\pi N) = \cos\left(2\pi n + \frac{5 \pi}{4}\right) = \cos\left(\frac{5 \pi}{4}\right) = -\frac{1}{\sqrt{2}}$

Here is a similar proof:

Note that for any integer n, cos(2npi+x)=cosx