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Castiel
 one year ago
Sum of a series question
Castiel
 one year ago
Sum of a series question

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Castiel
 one year ago
Best ResponseYou've already chosen the best response.1Okay I understand every part except the underlined. How do they get that? I can't seem to find what cancels with what here.

Castiel
 one year ago
Best ResponseYou've already chosen the best response.1I 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2familiar with telescoping ?

Castiel
 one year ago
Best ResponseYou've already chosen the best response.1not very unfortunately

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2*corrected Look at below pattern : \[\sum\limits_{n=2}^{\infty} [f(n)f(n+1)]\] can you expand it out ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2if 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

Castiel
 one year ago
Best ResponseYou've already chosen the best response.1Yeah I see, [( f(2)f(3) + f(3) f(4).... f(n) f(n+1)] and we get f(2)  f(n+1)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2you should get \[f(2)  \lim\limits_{n\to\infty} f(n+1)\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2the magic happens when that limit becomes 0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2lets try and use the same trick in current problem

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2start by splitting the middle term : \[\begin{align} &\dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left(\dfrac{1}{n1}\color{blue}{\dfrac{2}{n}}+\dfrac{1}{n+1}\right)\\~\\ &=\dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left(\dfrac{1}{n1}\color{blue}{\dfrac{1}{n}\dfrac{1}{n}}+\dfrac{1}{n+1}\right)\\~\\ \end{align}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2group first two terms group last two terms

Castiel
 one year ago
Best ResponseYou've already chosen the best response.1Omg it took me a while to understand the concept but I get it now: \[\frac{ 1 }{ 2 } \sum_{2}^{\infty}\left( \frac{ 1 }{ (n1)n } \frac{ 1 }{ \left( n+1 \right)n }\right)\] 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2start by splitting the middle term : \[\begin{align} &\dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left(\dfrac{1}{n1}\color{blue}{\dfrac{2}{n}}+\dfrac{1}{n+1}\right)\\~\\ &=\dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left(\dfrac{1}{n1}\color{blue}{\dfrac{1}{n}\dfrac{1}{n}}+\dfrac{1}{n+1}\right)\\~\\ &=\dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left(\dfrac{1}{n1}\color{blue}{\dfrac{1}{n}}\right)  \dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left( \color{blue}{\dfrac{1}{n}}\dfrac{1}{n+1}\right)\\~\\ \end{align}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2start by splitting the middle term : \[\begin{align} &\dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left(\dfrac{1}{n1}\color{blue}{\dfrac{2}{n}}+\dfrac{1}{n+1}\right)\\~\\ &=\dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left(\dfrac{1}{n1}\color{blue}{\dfrac{1}{n}\dfrac{1}{n}}+\dfrac{1}{n+1}\right)\\~\\ &=\dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left(\dfrac{1}{n1}\color{blue}{\dfrac{1}{n}}\right)  \dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left( \color{blue}{\dfrac{1}{n}}\dfrac{1}{n+1}\right)\\~\\ &=\dfrac{1}{2}\left(\dfrac{1}{21}\color{blue}{\lim\limits_{n\to\infty}\dfrac{1}{n}}\right)  \dfrac{1}{2}\left( \color{blue}{\dfrac{1}{2}}\lim\limits_{n\to\infty}\dfrac{1}{n+1}\right)\\~\\ \end{align}\]
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