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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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I really can't understand what is the identity the you want to prove. Could you try to typeset it (better if in LaTeX), with all the necessary parentheses, please? – Andrea Orta Sep 18 '12 at 18:25
Are you sure you have the right expression? For $n = 2$ you get: $\frac{1}{2 \cdot 6} = \frac{1}{2} - \frac{1}{10}$ which is clearly untrue. – Jakub Konieczny Sep 18 '12 at 18:25
If $n=1$ you are dividing by $0$. Also please check your parentheses: one extra left on the left side, it is not clear whether $(2n+2)$ should be in the numerator, and you must mean $1/(4n+2)$, not $(1/4)n+2$ or $1/(4n)+2$ on the right. \frac is your friend in this regard. – Ross Millikan Sep 18 '12 at 18:25
Also, unless there was supposed to be some sum involved, there is no need for induction. Just multiply everything out so as to get rid of fractions ;) – Jakub Konieczny Sep 18 '12 at 18:27
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. – Ross Millikan Sep 18 '12 at 18:37

4 Answers 4

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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@Peter Telescopy can be viewed as a form of induction specifically tailored for computing certain sums and products. – Bill Dubuque Sep 18 '12 at 19:00
Yes, I was just being silly, Bill! – Pedro Tamaroff Sep 18 '12 at 19:03

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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While this is the easiest approach, is not solving the problem the way the problem asks ;) – N. S. Sep 18 '12 at 18:43
@N.S.: I think it is a step toward the induction. – Ross Millikan Sep 18 '12 at 18:45
@RossMilikan he already guessed the right result, so he has a very easy induction problem, where he doesn't really need telescoping. – N. S. Sep 18 '12 at 18:49
@N.S. Telescopy does use induction - it's simply encapsulated (as a lemma) in the general closed form for a telescopic sum (whose inductive proof is easier in the abstract than in any specific case). So the natural way to prove this is to prove by induction the closed form for a telescopic sum, then specialize it to the case at hand - see my answer. Then, later, one can reuse the telescopy lemma to quickly tackle all such problems. – Bill Dubuque Sep 18 '12 at 19:09
@BillDubuque True, and this is the easiest way for us... But the problem asks the student to conjecture a closed formula, and prove it by induction. If you telescope it, you already get the answer... Is like the problem of calculating $\int_0^1 x^2 dx$... We typically expect the students to solve it the easy way, but if the Question explicitly asks to evaluate it using a Riemann sum, the easy solution is not good anymore... – N. S. Sep 18 '12 at 20:53

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.

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