I have the following practice problem from my Mathematics 2 course at EE university:
Do given equalities hold:
$$\sum_{n=1}^{+\infty}{\frac{1}{(n+1)^2}} = \sum_{n=1}^{+\infty}{\frac{n+1-n}{(n+1)^2}}=\sum_{n=1}^{+\infty}{\frac{1}{n+1}}-\sum_{n=1}^{+\infty}{\frac{n}{(n+1)^2}}$$
First equality is obviously true because it's a simple algebraic manipulation of adding a zero under the sigma symbol. The second one is the one that bothers me. I've had multiple attempts at this but they have failed miserably… I can see that on the right hand side I have two divergent series; both sequences $a_n = \frac{1}{n+1}$ and $b_n = \frac{n}{(n+1)^2}$ are asymptotically equivalent to $\frac{1}{n}$, that is $a_n \sim b_n \sim \frac{1}{n}$. This makes series $\sum_{n=1}^{+\infty}{a_n}$ and $\sum_{n=1}^{+\infty}{b_n}$ equiconvergent $\sum_{n=1}^{+\infty}{\frac{1}n}$, and since it theverges, both of them diverge as well. I know that the difference of two divergent series can be both convergent and divergent. With that in mind, the equality could hold.
I've tried introducing the limit definition of an infinite series into the right hand side but this was as far as I've made it. We have:
$$\sum_{n=1}^{+\infty}{\frac{1}{n+1}}-\sum_{n=1}^{+\infty}{\frac{n}{(n+1)^2}} = \lim_{m \to +\infty}\sum_{n=1}^m{\frac{1}{n+1}} – \lim_{m \to +\infty}\sum_{n=1}^m{\frac{n}{(n+1)^2}}$$
The reason why I do not know how to procced is that I am not aware of any theorems that would let me bring two divergent sequences into a common limit. I am basically clueless whether I am allowed to say that for two divergent sequences $a_n$ and $b_n$ say that $$\lim_{n \to +\infty}a_n \pm \lim_{n \to +\infty}b_n = \lim_{n \to +\infty}(a_n+b_n)$$. If this was the case the rest of the proof would be pretty easy. I can just bring the two fraction to a common denominator, cancel the terms and I would get the first term from the original question. If I cannot do this, I am pretty much clueless on how to approach this in any other way. Would proof by contradiction suffice? Should I try some series tests? Note that at this point in the practice material only Integral test and 4 basic comparison tests have been introduced. Can someone explain how this is supposed to be done? Thanks in advande!
Best Answer
The equality$$\sum_{n=1}^{+\infty}{\frac{n+1-n}{(n+1)^2}}=\sum_{n=1}^{+\infty}\frac1{n+1}-\sum_{n=1}^{+\infty}\frac n{(n+1)^2}\tag1$$neither holds nor doesn't hold. It is meaningless. In fact, neither$$\sum_{n=1}^{+\infty}{\frac{1}{n+1}}\quad\text{nor}\quad\sum_{n=1}^{+\infty}{\frac{n}{(n+1)^2}}$$are numbers. So, what's the meaning of the RHS of $(1)$? Even if you see them as $\infty$ (which makes sense), then $(1)$ will still be meaningless, since $\infty-\infty$ is undefined.