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

  1. As $n$ increases, $\frac{1}{n} - \frac{1}{n+1} \to 0$

  2. 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?

  • $\begingroup$ But by that logic, wouldn't it be true that $\lim_{n \to \infty} \frac{-(n+1)^{-1} + \sum_{k \geq n+1} \frac{1}{2^k k(k+1)}}{-n^{-1}} =1 $ which I guess is what I'm trying to show in the first place? As the sum goes to $0$ significantly faster, and $\frac{n+1}{n} \to 1$? $\endgroup$
    – Xiaomi
    Oct 7 '18 at 9:29
  • 1
    $\begingroup$ That's indeed correct, but also trivial. It is like writing $2^n+\sum\limits_{k\geqslant n}\dfrac1{k^2}\sim2^n\ (n→∞)$, i.e. overly dominating a small sum. $\endgroup$
    – Saad
    Oct 7 '18 at 9:34

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

Therefore your reasoning is correct as,

$$\lim_{n \to \infty} \frac{\sum_{k \geq n+1} \frac{1}{2^k k(k+1)}}{\frac{-1}{n}} = 0$$


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

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.


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