# Proof Question: Using Mathematical Induction

by Lococard
Tags: induction, mathematical, proof
 P: 25 1. The problem statement, all variables and given/known data Prove that, for all integers $n =>1$ $$\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}... + \frac{1}{n+1} = 1-\frac{1}{n+1}$$ 2. Relevant equations I am a little stuck on this question. :| 3. The attempt at a solution
 HW Helper P: 3,353 Well, the LHS can be expressed as $$\sum_{k=1}^n \frac{1}{k(k+1)}$$. Use partial fractions on that expression, you should get a telescoping series.
 P: 25 Anyone got any ideas with this problem? I cant find any notes regarding a similar question.
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PF Gold
P: 38,896

## Proof Question: Using Mathematical Induction

What was wrong with Gib Z's suggestion?
 P: 25 We haven't been taught that. Im a little unsure what he means as well. Could you possibly give more help? Sorry if im a pain.
P: 1,142
 Quote by Lococard We haven't been taught that. Im a little unsure what he means as well. Could you possibly give more help? Sorry if im a pain.
Putting it in partial fractions gives.

$\frac{1}{k(k+1)}=\frac{1}{k}\cdot\frac{1}{(k+1)}$ or $\equiv$?

Do you mean you don't understand summation?

All that expression is is your first expression for any given value of n. It's essentially the same thing.
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P: 26,167
 Quote by Lococard Prove that, for all integers $n =>1$ $$\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}... + \frac{1}{n+1} = 1-\frac{1}{n+1}$$
Hi Lococard!

For induction proofs, it often helps to give a name to the sum (if the question hasn't already doe so).

In this case, use the name a_n+1:

$$a_{n+1}\,=\,\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}... + \frac{1}{n+1}$$

Then what you have to prove is that, assuming that:
a_n = 1 - 1/n,
then:
a_n+1 = 1 - 1/(n+1).

You see how that makes the problem easier?
 P: 25 How would i put it into a partial fraction?
P: 1,142
 Quote by Lococard How would i put it into a partial fraction?
Well I gave one partial fraction that is in fact equivalent to your original equation. Put that equation into your calculator, plug in some numbers and then use the original equation, if they are equivalent it follows that both equations are different versions of the same partial fraction. Of course you could do this algebraically too, but I'll leave that to you. Tiny-tim has given you a nice hint there.
 P: 25 I've done them both on paper and i got the same result. Do i then add a_n+1 to the equation and get the formula in the lowest form?
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P: 26,167
 Quote by Lococard I've done them both on paper and i got the same result. Do i then add a_n+1 to the equation and get the formula in the lowest form?
Hi Lococard!

I'm not sure what you mean.

Just say a_n+1 - a_n = 1/n(n+1);

then, from the induction assumption, a_n = 1 - 1/n = (n - 1)/n.

So a_n+1 = a_n + 1/n(n+1) = (n - 1)/n + 1/n(n+1) = (n^2 - 1)/n(n+1) + 1/n(n+1) = … ?

Is that what you meant?
 P: 25 Alright, i got some more info on the question. It seems that: $$\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}... + \frac{1}{n+1} = 1-\frac{1}{n+1}$$ But. The number after $$\frac{1}{n+1}$$ would be $$\frac{(n+1)}{(n+1)(n+2)}$$ So. $$1 - \frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)}$$ = 1 - ?? Would would i simplify the rest of it?
 Sci Advisor HW Helper Thanks P: 26,167 ah! … you have a misprint … it should be $$\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}... + \frac{1}{n*(n+1)}\,=\,1-\frac{1}{n+1}\,.$$ Does that look better?
 P: 25 Oh whoops. Yeah thats what i meant
 P: 25 How would i do the next part?
 Sci Advisor HW Helper Thanks P: 26,167 $$1 - \frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)}$$ = $$\frac{n+1}{(n+1)} - \frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)}$$ = … ? (btw, wouldn't it have been easier to work out if you'd "gone down one", and writen: $$1 - \frac{1}{n} + \frac{1}{n(n+1)}$$ ?)
 P: 25 did you just substitute the 1 for a $\frac{n+1}{n+1}$? Im really stuck on simplifying the RHS.
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P: 26,167
 Quote by Lococard did you just substitute the 1 for a $\frac{n+1}{n+1}$?
Yes!

Tell me what bothers you about that …

Anyway, the next step is to use

$$\frac{n+1}{(n+1)} - \frac{1}{(n+1)}\,=\,\frac{n}{(n+1)}\,.$$

And then … ?

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