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Question: What would be the result of: $$\sum_{k=1}^{n}\frac{1}{n(n+2)}$$

My Approach:

Let $T_n$ denote the $n^{th}$ term of the given series. Then we have

$$T_1=\frac12 \left(\frac11-\frac13\right)$$ $$T_2=\frac12 \left(\frac12-\frac14\right)$$ $$T_3=\frac12 \left(\frac13-\frac15\right)$$

And so on up till $$T_n=\frac12 \left(\frac1n-\frac1{n+2}\right)$$

I can see that the series telescopes and the terms start to cut each other after an interval of one. My only problem is, how do I find the terms that remain in the end?

General Kenobi
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  • Compute the first partial sums or think about the terms that are never cancelled. –  Jul 07 '20 at 12:19
  • @YvesDaoust Sorry, but I don't know what you mean. Could you explain it better if you can? – General Kenobi Jul 07 '20 at 12:20
  • $(1-\color{red}{\frac{1}{3}})+(\color{blue}{\frac{1}{2}}-\color{green}{\frac{1}{4}})+(\color{red}{\frac{1}{3}}-\color{orange}{\frac{1}{5}})+(\color{green}{\frac{1}{4}}-...)+\dots$ The red terms cancel... the green terms cancel.. the orange terms cancel... etc... there will be some terms which didn't cancel at the very front and some which didn't cancel at the very end... – JMoravitz Jul 07 '20 at 12:26

1 Answers1

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For all $k \geq 1$ you have $$ \frac{1}{k(k+2)} = \frac{1}{2} \left( \frac{1}{k} - \frac{1}{k+2} \right) $$ so $$ \begin{aligned} \sum_{k=1}^n \frac{1}{k(k+2)} &= \frac{1}{2}\sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+2} \right) \\ &= \frac{1}{2}\left( \sum_{k=1}^n \frac{1}{k} - \sum_{k=1}^n \frac{1}{k+2} \right) \\ &= \frac{1}{2}\left( \sum_{k=1}^n \frac{1}{k} - \sum_{k=3}^{n+2} \frac{1}{k} \right) \\ &= \frac{1}{2}\left( 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right) \\ \end{aligned} $$

DodoDuQuercy
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  • How did you find out that $-\frac1{n-1}$ remains in the end? – General Kenobi Jul 07 '20 at 12:29
  • You can see that the only $+\frac{1}{k}$ remaining are for $k \in { 1,2 }$ and that the only $-\frac{1}{k}$ remaining are for $k \in { n+1, n+2 }$. All the other terms cancel out since they appear in both sums – DodoDuQuercy Jul 07 '20 at 12:30