anonymous
  • anonymous
calculus exam tomorrow please help me, find the sum of the series :
Mathematics
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[ \sum_{k=1}^{\infty}1/k(k+3)\]
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align}& \sum_{1}^{\infty} \dfrac{1}{k(k+3)}\hspace{.33em}\\~\\ & \implies \sum_{1}^{\infty} \dfrac{3}{3k(k+3)}\hspace{.33em}\\~\\ & \implies \sum_{1}^{\infty} \dfrac{k+3-k}{3k(k+3)}\hspace{.33em}\\~\\ & \implies \sum_{1}^{\infty} \dfrac{1}{3k}-\dfrac{1}{3(k+3)}\hspace{.33em}\\~\\ & \implies \dfrac{1}{3}\sum_{1}^{\infty} \dfrac{1}{k}-\dfrac{1}{(k+3)}\hspace{.33em}\\~\\ \end{align}}\) now try to put some values up to 9 or 10 and see if it gives telescooping series
mathmath333
  • mathmath333
https://en.wikipedia.org/wiki/Telescoping_series

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anonymous
  • anonymous
@mathmath333 thank you so much..

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