Is it allowed to separate infinite sums?

Question

Just a quick question, is it allowed to separate infinite sums. For example I want to prove that $\sum \frac{n-1}{n^2}$ diverges as follows: $$ \sum \frac{n-1}{n^2} = \sum \left( \frac{n}{n^2} - \frac{1}{n^2} \right) = \sum \frac{1}{n} - \sum \frac{1}{n^2} \geq \sum \frac{1}{n} -2 = +\infty, $$ Since $\sum \frac{1}{n^2} \leq 2$ and $\sum \frac{1}{n}$ diverges. Is this valid? I'm not sure because of the infinity of the sums, for finite sums this is allowed of course.

Answer 1

Someone correct me if I am wrong.

If we know the behaviors of $\sum a_n$ and $\sum b_n$ and want to figure out the behavior of $\sum (a_n + b_n)$ from them...

If you have $\sum (a_n + b_n)$ and $\sum a_n$ and $\sum b_n$ both converge then $\sum a_n + \sum b_n = \sum (a_n + b_n)$ and so it is allowed.

If you have $\sum (a_n + b_n)$ and $\sum a_n$ diverges but $\sum b_n$ converges we can conclude by contradiction that $\sum (a_n + b_n)$ diverges but NOT because we can separate $\sum(a_n + b_n) = \sum a_n + b_n$-- we can NOT do that--- but because IF $\sum(a_n + b_n)$ converged we would be able to separate $\sum a_n + \sum b_n$ and that sum would converge, which it does not so that is a contradiction. So we can conclude $\sum(a_n + b_n) $ does not converge.

And if we have $\sum (a_n + b_n)$ and $\sum a_n$ and $\sum b_n$ fail to converge then ... we got nothing. There's nothing we can conclude.

At least I think that is correct.

If on the other hand we know the behavoir of $\sum (a_n + b_n)$ and the behavior of $\sum b_n$ and want to figure out the behavior of $\sum a_n$ we can do similar:.

If $\sum (a_n + b_n)$ converges and $\sum b_n$ converge then $\sum (a_n + b_n) - \sum b_n = \sum (a_n + b_n - b_n) = \sum a_n $ converges is allowed.

If $\sum (a_n + b_n)$ diverges and $\sum b_n$ converges we can assume by contradiction that $a_n$ doesn't converge. If it did $\sum a_n + \sum b_n = \sum (a_n + b_n)$ would converge but it doesn't.

If $\sum (a_n + b_n)$ converges but $\sum b_n$ does not then by contradiction $a_n$ does not. If is did we could have $\sum (a_b + b_n) - \sum a_n = \sum (a_b + b_n - a_n) = \sum b_n$ converges. But it doesn't.

And finaly if $\sum (a_n + b_n)$ diverges and $\sum b_n$ diverges we can't do or conclude anything.