Question

calculus exam tomorrow
please help me,
find the sum of the series :

Answer 1

\[ \sum_{k=1}^{\infty}1/k(k+3)\]

Answer 2

\(\large \color{black}{\begin{align}& \sum_{1}^{\infty} \dfrac{1}{k(k+3)}\hspace{.33em}\\~\\
& \implies \sum_{1}^{\infty} \dfrac{3}{3k(k+3)}\hspace{.33em}\\~\\
& \implies \sum_{1}^{\infty} \dfrac{k+3-k}{3k(k+3)}\hspace{.33em}\\~\\
& \implies \sum_{1}^{\infty} \dfrac{1}{3k}-\dfrac{1}{3(k+3)}\hspace{.33em}\\~\\
& \implies \dfrac{1}{3}\sum_{1}^{\infty} \dfrac{1}{k}-\dfrac{1}{(k+3)}\hspace{.33em}\\~\\
\end{align}}\)


now try to put some values up to 9 or 10 and see if it gives telescooping series

Answer 3

https://en.wikipedia.org/wiki/Telescoping_series

Answer 4

@mathmath333 thank you so much..