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## Homework Statement

For each positive integer n, let [tex]S_{n} = \frac{1}{n(n+1)} + \frac{1}{(n+1)(n+2)} + ... + \frac{1}{(2n-1)2n}.[/tex]

(a) Calculate [tex]S_{1}, S_{2}, S_{3}[/tex]. Then use this data to guess a simple formula for [tex]S_{n}[/tex].

(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 [tex]\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}[/tex] 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.

## Homework Equations

Result 6.6: For every positive integer n:

[tex]\frac{1}{(2)(3)} + \frac{1}{(3)(4)} + ... + \frac{1}{(n+1)(n+2)} = \frac{n}{2n+4}.[/tex]

## The Attempt at a Solution

Part a)

Since [tex]S_{1} = \frac{1}{2}, S_{2} = \frac{1}{4}, S_{3} = \frac{1}{6}[/tex], a good guess would be that [tex]S_{n} = \frac{1}{2n}[/tex].

Part b)

For n=1, [tex]\frac{1}{(2n-1)2n}=\frac{1}{(2(1)-1)2(1)}=\frac{1}{2}[/tex], so the entire sum is given by [tex]\frac{1}{n(n+1)} =\frac{1}{(1)((1)+1)} = \frac{1}{2}[/tex]. Thus, [tex]S_{n} = \frac{1}{2n}[/tex] is true for n=1. By induction, let k be an arbitrary integer and assume that [tex]\frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} + ... + \frac{1}{(2k-1)2k} = \frac{1}{2k}[/tex]. We want to prove that [tex]\frac{1}{(k+1)((k+1)+1)} + \frac{1}{((k+1)+1)((k+1)+2)} + ... + \frac{1}{(2(k+1)-1)(2(k+1))} = \frac{1}{2(k+1)}[/tex], which when simplified gives: [tex]\frac{1}{(k+1)(k+2)} + \frac{1}{(k+2)(k+3)} + ... + \frac{1}{(2k+1)(2k+2)} = \frac{1}{2(k+1)}[/tex].

I don't know where to go from here on (b) and I have no idea how Result 6.6 could help prove the result in a different way, and I'm completely lost on (d) as well. If anyone could help me out that would be massively appreciated, I've been spending hours trying to figure this out. I've run into tons of dead ends and what I've typed up here is the only thing that I know is for sure correct.

Thanks!