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I just do not understand how to find the sum of this infinite series; I understand that through partial fraction composition you come out with a sum of that ends with $ \frac{1}{n} - \frac{1}{n+3} $ but I do not know how to apply this to find the sum of this series.

$$ \sum_{i=3}^\infty\frac{3}{n(n+3)} $$

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The operating principle here is called telescoping. Think of an old fashioned shipboard (pirate) spyglass-style telescope that folds inward to become shorter.

You already know how to decompose into partial fractions. Let's write down the first few terms of the series formed after that decomposition:

$\frac 13 - \frac 16 + \frac 14 - \frac 17 + \frac 15 - \frac 18 + \frac 16 - \frac 19 + \frac 17 - \frac 1{10} + ...$

Note how, starting from $\frac 16$ onward every initial negative term has a positive counterpart later on. In essence, all negative terms will simply vanish by cancellation, as will all positive terms after $\frac 15$. So your entire series "telescopes" to become the really short $\frac 13 + \frac 14 + \frac 15 = \frac{47}{60}$.

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Hint: this is a telescopic series: $$ \sum_{n=3}^\infty\left[\frac1n-\frac{1}{n+3}\right]=\frac13+\frac14+\frac15, $$ in which only three first positive terms survive, the other being cancelled by the preceeding negative ones.

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  • $\begingroup$ The series sum starts from index $3$ not $1$. $\endgroup$ – Deepak Jan 26 at 0:48
  • $\begingroup$ @Deepak Thanks for noting this. $\endgroup$ – user Jan 26 at 0:50
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Hint: $$\frac3{n(n+3}=\frac 1n-\frac 1{n+3}$$ Write out the beginning of this sum, and you should find telescoping terms.

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This is a special case of the following theorem:

Given $k$ reals $(a_k)_{j=1}^k$, the sum $\sum_{n=1}^{\infty} \sum_{j=1}^k \dfrac{a_j}{(n-1)k+j} $ converges if and only if $\sum_{j=1}^k a_j= 0$.

This problem is the case $k=4, a_1=1, a_2=0, a_3 = 0, a_4 = -1$. Therefore the sum converges.

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    $\begingroup$ The problem appears to be to find the value of the sum, not whether it converges. $\endgroup$ – 6005 Jan 26 at 2:49
  • $\begingroup$ Yep. I posted too fast. I think I'll make my answer into a question with a proof. Thanks. $\endgroup$ – marty cohen Jan 26 at 5:19

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