Determine the convergence or divergence of f(n) = 1 + (n / (n+1)) cos (nπ / 2)

Consider the function $f(n)$ defined by

$f(n) = 1 + \left( \frac{n}{n+1} \right) \cos \left( \frac{n \pi}{2} \right).$

Determine whether the sequence $\{ f(n) \}$ converges or diverges, and if it converges find the limit.

The sequence $\{ f(n) \}$ diverges.
Proof. Suppose otherwise. Then there exists a real number $L$ and a positive integer $N$ such that

$|f(n) - L| < \varepsilon \qquad \text{for all } \varepsilon > 0 \quad \text{for all } n > N.$

Since $N$ is positive, we know $4N > N$ and $4N+2 > N$ . Letting $\varepsilon = \frac{1}{2}$ we then have

$|f(4N) - L| < \frac{1}{2} \qquad \text{and} \qquad |f(4N+2) - L| < \frac{1}{2},$

which implies

$\left| 1 + \left( \frac{4N}{4N+1} \right) \cos (2N \pi) - L \right| < \frac{1}{2} \qquad \text{and} \qquad \left| 1 + \left( \frac{4N+2}{4N+3}\right) \cos (2N \pi + \pi) - L \right| < \frac{1}{2}.$

Therefore,

$\left| 1 + \frac{4N}{4N+1} - L \right| < \frac{1}{2} \quad \text{and} \quad \left| L - 1 + \frac{4N+2}{4N+3} \right| < \frac{1}{2}.$

So, adding these together and using the triangle inequality we then have,

$\begin{align*} &&\left| 1 + \frac{4N}{4N+1} - L \right| + \left| L - 1 + \frac{4N+2}{4N+3} \right| &< 1 \\[9pt] \implies && \left| \frac{4N}{4N+1} + \frac{4N+2}{4N+3} \right| &< 1 \\[9pt] \implies && \frac{4N}{4N+1} + \frac{4N+2}{4N+3} &< 1. \end{align*}$

But, both $\frac{4N}{4N+1}$ and $\frac{4N+2}{4N+3}$ are greater than $\frac{3}{4}$ for all positive integers $N$ , contradicting that $\frac{4N}{4N+1} + \frac{4N+2}{4N+3} < 1$ for some positive integer $N$ . Hence, this sequence can have no such limit $L$ and must diverge $. \qquad \blacksquare$