I'm working on a very lengthy proof and the final step to complete it is for me to show that
$$\frac{-1}{n+1} + \sum_{k \geq n+1} \frac{1}{2^k k(k+1)} \sim \frac{-1}{n}$$
With the approximation becoming increasingly accurate as $n \to \infty$.
My justification for this is that
As $n$ increases, $\frac{1}{n} - \frac{1}{n+1} \to 0$
As $n$ increases, $\sum_{k \geq n+1} \frac{1}{2^k k(k+1)} \to 0$ faster than $\frac{1}{n}-\frac{1}{n+1} \to 0$
Hence the approximation seems justifiable to me.
I'm not good with series and convergence, and my argument here seems very informal (and possibly wrong?). Could someone help explain a better way to justify this approximation, or explain how it might be wrong if that's the case?
$$\lim_{n \to \infty} \frac{\frac{-1}{(n+1)} + \sum_{k \geq n+1} \frac{1}{2^k k(k+1)}}{\frac{-1}{n}} =1$$
Therefore your reasoning is correct as,
$$\lim_{n \to \infty} \frac{\sum_{k \geq n+1} \frac{1}{2^k k(k+1)}}{\frac{-1}{n}} = 0$$
And,
$$\lim_{n \to \infty} \frac{\frac{-1}{(n+1)}}{\frac{-1}{n}} = \lim_{n \to \infty}\frac{n}{n+1}= 1$$
Sum of both is 1.
I'm not sure why this is so necessary for your proof as if n tends to infinity, the whole limit just becomes 0. In cases where n is not infinity, the relationship between the 2 isn't very strong.