Finding the limit of $\lim_{n\rightarrow\infty}(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+...[\text{upto n times}])$

Question

The question is as follows-

Evaluate the limit.

$$\lim_{n\rightarrow\infty}(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+...[\text{upto n times}])$$

I have no idea on how to solve this limit.

Thanks for any help,response or hint!!

Answer 1

Use formula: $$\frac{1}{n^3-n}=\frac1{(n-1)n(n+1)}=\frac12 \left(\frac1{n(n-1)}-\frac1{n(n+1)} \right)$$

Then $$\lim_{n\rightarrow\infty}\left(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+...+\frac{1}{n^3-n}\right)=$$ $$=\lim_{n\rightarrow\infty}\frac12\left(\frac{1}{2\cdot1}-\frac{1}{2\cdot3}+\frac{1}{3\cdot2}-\frac{1}{3\cdot4}+\frac{1}{4\cdot3}-...-\frac{1}{n(n+1}\right)=$$

$$=\lim_{n\rightarrow\infty}\frac12\left(\frac12-\frac{1}{n(n+1)} \right)=\frac14$$

Answer 2

HINT:

$$\dfrac1{n(n-1)(n+1)}=\dfrac{n+1-n}{n(n-1)(n+1)}=\dfrac1{n(n-1)}-\dfrac1{(n-1)(n+1)}$$

$$\dfrac1{n(n-1)}=\dfrac{n-(n-1)}{n(n-1)}=f(n-1)-f(n)$$

where $f(m)=\dfrac1m$

$$\dfrac2{(n-1)(n+1)}=\dfrac{n+1-(n-1)}{(n-1)(n+1)}=f(n-1)-f(n+1)$$

$$2\sum_{r=2}^n\dfrac1{r^3-r}=2\sum_{r=2}^n\left\{f(r-1)-f(r)\right\}-\sum_{r=2}^n\left\{f(r-1)-f(r+1)\right\}$$

Set a few values of $n$ to recognize the Telescoping Series.

Answer 3

$$2\sum_2^{n+1}\frac{1}{(m-1)m(m+1)}=\sum_2^{n+1}\left(\frac{1}{m(m-1)}-\frac{1}{(m+1)m}\right)=\frac{1}{2}-\frac{1}{(n+1)(n+2)}$$ So the limit is $\frac{1}{4}$.