Calculate $\lim_{n\to\infty} (\frac{1}{{1\cdot2}} + \frac{1}{{2\cdot3}} + \frac{1}{{3\cdot4}} + \cdots + \frac{1}{{n(n + 1)}})$

Question

Calculate

$$\lim_{n\to\infty} \left(\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \cdots + \frac{1}{n(n + 1)}\right). $$

If reduce to a common denominator we get

$$\lim _{n\to\infty}\left(\frac{X}{{n!(n + 1)}}\right).$$

How can I find $X$ and calculate limit?

Answer 1

The limit you want to have calculated is the following $$\sum_{k=1}^{\infty}\frac{1}{k(k+1)}=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k(k+1)}.$$ Note that $\tfrac{1}{k(k+1)}=\tfrac{1}{k}-\tfrac{1}{k+1}$, so this is a telescoping series, meaning that consecutive terms cancel eachother out. It is then easy to see that for all $n$ we have $$\sum_{k=1}^n\frac{1}{k(k+1)}=1-\frac{1}{n+1}.$$ Now the limit is easy to evaluate; we find that $$\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k(k+1)}=\lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)=1.$$

Answer 2

This won't get you far. Try $\frac{1}{n(n + 1)} = \frac{n + 1 - n}{n(n+1)} = \frac{1}{n} - \frac{1}{n + 1}$ instead.

Answer 3

We can write sum as $\displaystyle \lim_{n\rightarrow \infty}\sum_{r=1}^{n}\left[\frac{1}{r\cdot (r+1)}\right] = \lim_{n\rightarrow \infty}\sum_{r=1}^{n}\left[\frac{1}{r}-\frac{1}{r+1}\right]$

Now Using Telescopin Sum(or Simply open Sigma)

We get Sum $\displaystyle = \lim_{n\rightarrow \infty}\left(1-\frac{1}{n+1}\right) = 1-0 =1$