Question

Sum of a series question

Answer 1

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

Answer 2

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.

Answer 3

familiar with telescoping ?

Answer 4

not very unfortunately

Answer 5

*corrected
Look at below pattern :
\[\sum\limits_{n=2}^{\infty} [f(n)-f(n+1)]\]

can you expand it out ?

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

the magic happens when that limit becomes 0

Answer 10

lets try and use the same trick in current problem

Answer 11

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}\]

Answer 12

group first two terms
group last two terms

Answer 13

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

Answer 14

Thank you!

Answer 15

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}\]

Answer 16

or like that too :)

Answer 17

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}\]