Study the convergence of this succession

Question

I need help with this exercise.

$$ U_n=\frac{n \times cos(n \pi)}{2n+1} $$

What I've done so far is recognize the main difficulty of the problem which is the $cos(n\pi)$, an alternating term due to the fact that it can only go until number $1$ and $-1$ back again.

$$ U_n=\frac{n}{2n+1}\times cos(n \pi) $$

Writing the exercise like this, gives a little perspective of what it should be done to know if the succession does really converge.

My problem is that I don't know if I should do the limit of the non-alternating term (to figure out if it converges or not) or if I should just finish the exercise by saying that, because of the $cos(n\pi)$, the succession has a limit and, therefore, it converges.

Thank you so much If someone could help I would really appreciate it

Answer 1

HINT

Let consider the subsequences for

• $n=2k \implies U_{2k}=\frac{2k \times cos(2k \pi)}{4k+1}=\frac{2k}{4k+1}\to \frac12$
• $n=2k+1 \implies U_{2k+1}=\frac{(2k+1) \times cos(2k \pi+\pi)}{4k+2}=\frac{-(2k+1)}{4k+2}\to -\frac12$

and note that those subsequences tend to different limits.

Then remember that if limit exists all the subsequecens tend to the same limit (i.e. it is a necessary condition for the existence of the limit).