## Castiel one year ago Sum of a series question

1. Castiel

Okay I understand every part except the underlined. How do they get that? I can't seem to find what cancels with what here.

2. Castiel

I have a couple more that are solved pretty much in the same way but I can't seem to grasp what they are actually doing here.

3. ganeshie8

familiar with telescoping ?

4. Castiel

not very unfortunately

5. ganeshie8

*corrected Look at below pattern : $\sum\limits_{n=2}^{\infty} [f(n)-f(n+1)]$ can you expand it out ?

6. ganeshie8

if you expand, you will notice that everything cancels out except for the first term and last term maybe first do it and convince yourself that

7. Castiel

Yeah I see, [( f(2)-f(3) + f(3)- f(4).... f(n)- f(n+1)] and we get f(2) - f(n+1)

8. ganeshie8

you should get $f(2) - \lim\limits_{n\to\infty} f(n+1)$

9. ganeshie8

the magic happens when that limit becomes 0

10. ganeshie8

lets try and use the same trick in current problem

11. ganeshie8

start by splitting the middle term : \begin{align} &\dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left(\dfrac{1}{n-1}\color{blue}{-\dfrac{2}{n}}+\dfrac{1}{n+1}\right)\\~\\ &=\dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left(\dfrac{1}{n-1}\color{blue}{-\dfrac{1}{n}-\dfrac{1}{n}}+\dfrac{1}{n+1}\right)\\~\\ \end{align}

12. ganeshie8

group first two terms group last two terms

13. Castiel

Omg it took me a while to understand the concept but I get it now: $\frac{ 1 }{ 2 } \sum_{2}^{\infty}\left( \frac{ 1 }{ (n-1)n }- \frac{ 1 }{ \left( n+1 \right)n }\right)$ and then the same as before first and last term, the last term when goes to infinity is zero and we are left with 1/2 * 1/2

14. Castiel

Thank you!

15. ganeshie8

start by splitting the middle term : \begin{align} &\dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left(\dfrac{1}{n-1}\color{blue}{-\dfrac{2}{n}}+\dfrac{1}{n+1}\right)\\~\\ &=\dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left(\dfrac{1}{n-1}\color{blue}{-\dfrac{1}{n}-\dfrac{1}{n}}+\dfrac{1}{n+1}\right)\\~\\ &=\dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left(\dfrac{1}{n-1}\color{blue}{-\dfrac{1}{n}}\right) - \dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left( \color{blue}{\dfrac{1}{n}}-\dfrac{1}{n+1}\right)\\~\\ \end{align}

16. Castiel

or like that too :)

17. ganeshie8

start by splitting the middle term : \begin{align} &\dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left(\dfrac{1}{n-1}\color{blue}{-\dfrac{2}{n}}+\dfrac{1}{n+1}\right)\\~\\ &=\dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left(\dfrac{1}{n-1}\color{blue}{-\dfrac{1}{n}-\dfrac{1}{n}}+\dfrac{1}{n+1}\right)\\~\\ &=\dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left(\dfrac{1}{n-1}\color{blue}{-\dfrac{1}{n}}\right) - \dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left( \color{blue}{\dfrac{1}{n}}-\dfrac{1}{n+1}\right)\\~\\ &=\dfrac{1}{2}\left(\dfrac{1}{2-1}\color{blue}{-\lim\limits_{n\to\infty}\dfrac{1}{n}}\right) - \dfrac{1}{2}\left( \color{blue}{\dfrac{1}{2}}-\lim\limits_{n\to\infty}\dfrac{1}{n+1}\right)\\~\\ \end{align}