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Castiel

  • one year ago

Sum of a series question

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  1. Castiel
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    familiar with telescoping ?

  4. Castiel
    • one year ago
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    not very unfortunately

  5. ganeshie8
    • one year ago
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    *corrected Look at below pattern : \[\sum\limits_{n=2}^{\infty} [f(n)-f(n+1)]\] can you expand it out ?

  6. ganeshie8
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    you should get \[f(2) - \lim\limits_{n\to\infty} f(n+1)\]

  9. ganeshie8
    • one year ago
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    the magic happens when that limit becomes 0

  10. ganeshie8
    • one year ago
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    lets try and use the same trick in current problem

  11. ganeshie8
    • one year ago
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    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
    • one year ago
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    group first two terms group last two terms

  13. Castiel
    • one year ago
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    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
    • one year ago
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    Thank you!

  15. ganeshie8
    • one year ago
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    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
    • one year ago
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    or like that too :)

  17. ganeshie8
    • one year ago
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    start by splitting the middle term : \[\begin{align} &\dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left(\dfrac{1}{n-1}\color{blue}{-\dfrac{2}{n}}+\dfrac{1}{n+1}\right)\\~\\ &=\dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left(\dfrac{1}{n-1}\color{blue}{-\dfrac{1}{n}-\dfrac{1}{n}}+\dfrac{1}{n+1}\right)\\~\\ &=\dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left(\dfrac{1}{n-1}\color{blue}{-\dfrac{1}{n}}\right) - \dfrac{1}{2}\sum\limits_{n=2}^{\infty}\left( \color{blue}{\dfrac{1}{n}}-\dfrac{1}{n+1}\right)\\~\\ &=\dfrac{1}{2}\left(\dfrac{1}{2-1}\color{blue}{-\lim\limits_{n\to\infty}\dfrac{1}{n}}\right) - \dfrac{1}{2}\left( \color{blue}{\dfrac{1}{2}}-\lim\limits_{n\to\infty}\dfrac{1}{n+1}\right)\\~\\ \end{align}\]

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