# Proof of a summation by the use of a result of a different summation

For each positive integer $n$, let $$S_n= \frac1{n(n+1)} + \frac1{(n+1)(n+2)} +\dots + \frac1{(2n−1)2n}.$$

(a) Calculate $S_1,S_2,S_3$. Then use this data to guess a simple formula for $S_n$.
(b) Prove your guess in part (a) by mathematical induction.
(c) Use Result 6.6 on page 136 to give another proof of your guess.
(d) Prove that $$\frac1{k(k+1)}=\frac1k−\frac1{k+1}$$ for all positive real numbers $k$. Use this to give yet another proof of your guess in part (a). This method of proof is called telescoping.

Result 6.6: For every positive integer $n$: $$\frac1{(2)(3)} + \frac1{(3)(4)} +\dots + \frac1{(n+1)(n+2)} = \frac{n}{2n+4}.$$

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I've already given a proof of 2(b) for that $S_{n} = \frac1{2n}$, but I am stuck on (c) and (d).

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For (d), did you check that performing the subtraction on the right of your first expression yields the left side? – J. M. Oct 19 '11 at 4:57
For (c): replace the $n$ with $2n-2$ and with $n-2$ to yield two expressions you can subtract... – J. M. Oct 19 '11 at 5:00

HINT for (c): No matter what the terms $a_k$ are, $$a_{n+1}+a_{n+2}+\dots+a_{2n} = (a_3+a_4+\dots+a_{2n})-(a_3+a_4+\dots+a_n)\::$$ just look at what cancels out.
For (d), verifying that $$\frac1{k(k+1)}=\frac1k-\frac1{k+1}$$ is just algebra. Once you have that, \begin{align*} S_n &= \frac1{n(n+1)} + \frac1{(n+1)(n+2)} +\dots + \frac1{(2n−1)2n}\\ &= \left(\frac1n-\frac1{n+1}\right)+\left(\frac1{n+1}-\frac1{n+2}\right)+\dots+\left(\frac1{2n-1}-\frac1{2n}\right), \end{align*} which simplifies very easily once you clear out the parentheses.