Convergence of Sum of Sequences

Question

This week, I learned a bit more about limits, convergence and divergence. I was given a sum of two sequences and asked to tell whether or not it is convergent, and what its limit is:

$a_n := (-1)^n + \frac{1}{n^2 +1}$

which I re-wrote into

$\lim_{n\to \infty}(-1)^n +\lim_{n\to \infty}\frac{1}{n^2 +1}$

I noticed that $\lim_{n\to \infty}(-1)^n$ isn't convergent, whereas the latter is convergent and has the limit of $0$. That is why I'm not entirely sure whether $a_n$ is convergent or not, and got confused.

I hope someone can clear my doubts and explain their answer to me! Thank you.

Answer 1

Let $u_n=(-1)^n$ and $v_n=\frac{1}{n^2+1}$.

$(u_n)$ is divergent since $$\lim u_{2n}\ne \lim u_{2n+1}$$

$(v_n)$ is convergent since

$$\lim v_n=0.$$

the sum of a convergent sequence and a divergent one is DIVERGENT.

Answer 2

Let consider

• n odd $\implies a_n := (-1)^n + \frac{1}{n^2 +1} \to -1$

• n even $\implies a_n := (-1)^n + \frac{1}{n^2 +1} \to 1$

what can we conclude, recalling that for a convergent sequence all the subsequence need to converge to the same limit?

Answer 3

You should look at the global situation rather than focusing on the specific example.

The sum of a convergent sequence and a divergent one diverges. If this wasn't the case the difference of the sum and the convergent sequence would converge. That can't be as this is equal to the divergent sequence.