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anonymous
 one year ago
calculus exam tomorrow
please help me,
find the sum of the series :
anonymous
 one year ago
calculus exam tomorrow please help me, find the sum of the series :

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ \sum_{k=1}^{\infty}1/k(k+3)\]

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{\begin{align}& \sum_{1}^{\infty} \dfrac{1}{k(k+3)}\hspace{.33em}\\~\\ & \implies \sum_{1}^{\infty} \dfrac{3}{3k(k+3)}\hspace{.33em}\\~\\ & \implies \sum_{1}^{\infty} \dfrac{k+3k}{3k(k+3)}\hspace{.33em}\\~\\ & \implies \sum_{1}^{\infty} \dfrac{1}{3k}\dfrac{1}{3(k+3)}\hspace{.33em}\\~\\ & \implies \dfrac{1}{3}\sum_{1}^{\infty} \dfrac{1}{k}\dfrac{1}{(k+3)}\hspace{.33em}\\~\\ \end{align}}\) now try to put some values up to 9 or 10 and see if it gives telescooping series

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@mathmath333 thank you so much..
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