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I can prove the base case and get that $1/3=1/3$, but i can't get any further with the $n+1$ case. Can someone help me?

I was told to conjecture a formula for the sum $\frac{1}{3} + \frac{1}{{3 \cdot 5}} + \cdots + \frac{1}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}}$ I thought I figured out that this sum was equal to $\frac{1}{2} { - \frac{1}{{4n + 2}}} $ but I'm starting to think I'm wrong about that. After we have the formula, we were told to prove our conjecture using induction.

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    Based on your added comment, you want the left to be $\sum_{i=1}^n \frac 1{2i-1}\frac 1{2i+1}=\frac 12 - \frac 1{4n+2}$. Note the $1$'s, not $2$'s and the sum on the left.2012-09-18

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Hint $\ $ The sum telescopes since for $\rm\:f(k) = 1/(4k-2)\:$ we have

$\rm f(k+1)-f(k)\ =\ \frac{1}{4k+2} - \frac{1}{4k-2}\ =\ -\frac{1}{(2k-1)(2k+1)}$

so an inductive proof is a special case of the inductive proof of the closed form for a telescopic sum:

$\rm\sum_{k=1}^n f(k\!+\!1)-f(k) = - f(1) + \color{#C00}{f(2) - f(2) + \cdots + f(n)-f(n)} + f(n\!+\!1)\, =\, f(n\!+\!1)-f(1)$

But it is easy to turn the above ellipses into a rigorous inductive proof. Then your problem is simply a corollary of this general telescopy lemma, for the special value of $\rm\:f(k)\:$ given above.

The advantage of proving it this way is that - with no extra effort - you now have a general lemma that works to prove (by induction!) all summation identities of this form. Note that even though the induction has been abstracted out into a proof of a more general telescopy lemma, a proof invoking the telescopy lemma still counts as a proof by induction. The induction simply has been encapsulated in the proof of the lemma, which need not be repeated inline every time it is invoked. DivideAbstract and conquer - once you've seen one telescopic induction proof you've seen them all!

You can find more examples and further discussion in my prior posts on telescopy.

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    Yes, I was just being silly, Bill!2012-09-18
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What you want to prove, apparently, is that

$\sum_{k=1}^n \frac{1}{(2k-1)(2k+1)}=\frac 1 2 - \frac 1 {2(2n+1)}$

As you say, the base case, $n=1$ is true. Suppose true for $n$, and analyze $n+1$. We get

$\sum_{k=1}^{n+1} \frac{1}{(2k-1)(2k+1)}=\frac 1 2 - \frac 1 {2(2(n+1)+1)}$

$\sum_{k=1}^{n} \frac{1}{(2k-1)(2k+1)}+\frac{1}{(2(n+1)-1)(2(n+1)+1)}=\frac 1 2 - \frac 1 {2(2(n+1)+1)}$

By the inductive hypothesis, this is, $\frac 1 2 - \frac 1 {2(2n+1)}+\frac{1}{(2(n+1)-1)(2(n+1)+1)}=\frac 1 2 - \frac 1 {2(2(n+1)+1)}$

$ - \frac{1}{{2(2n + 1)}} + \frac{1}{{(2n + 1)(2n + 3)}} = - \frac{1}{{2(2n + 3)}}$

Multiply through ${(2n + 1)(2n + 3)}$ to get

$ - \frac{{(2n + 3)}}{2} + 1 = - \frac{{(2n + 1)}}{2}$

In the end, you get

$ - 2n - 3 + 2 = - 2n - 1$ $ - 2n -1 = - 2n - 1$

which is true, so the inductive step is complete, and the theorem is proven.

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Hint: This is a telescoping series. Note that $\frac 1{2i-1}\frac 1{2i+1}=\frac 12 \left(\frac 1{2i-1}-\frac 1{2i+1}\right)$ so neighboring terms cancel.

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    @N.S. But after one discovers the closed form, one can easily go further and discover that it works by telescopy, and immediately discover and prove general telescopy.2012-09-18
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Saw this type of problem before.

You need to find a closed for for

$\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5}+...+\frac{1}{(2n-1)(1n+2)} $

and then prove it by induction.

Your guess looks right, now you can do the induction:

$\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5}+...+\frac{1}{(2n-1)(1n+2)} =\frac{1}{2}-\frac{1}{2n+1} \,.$

Alternately, you can use partial fraction decomposition to find a simple formula for $\frac{1}{(2k-1)(2k+1)} \,,$ and then get a telescopic sum. There is no need to prove this partial fraction decomposition by induction, and this approach is not really a direct induction proof.