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Proof Question: Using Mathematical Induction 
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#1
Mar2408, 05:08 AM

P: 25

1. The problem statement, all variables and given/known data
Prove that, for all integers [itex]n =>1[/itex] [tex]\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}... + \frac{1}{n+1} = 1\frac{1}{n+1}[/tex] 2. Relevant equations I am a little stuck on this question. : 3. The attempt at a solution 


#2
Mar2408, 06:29 AM

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P: 3,348

Well, the LHS can be expressed as
[tex]\sum_{k=1}^n \frac{1}{k(k+1)}[/tex]. Use partial fractions on that expression, you should get a telescoping series. 


#3
Mar2508, 02:03 AM

P: 25

Anyone got any ideas with this problem?
I cant find any notes regarding a similar question. 


#4
Mar2508, 06:07 AM

Math
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PF Gold
P: 39,682

Proof Question: Using Mathematical Induction
What was wrong with Gib Z's suggestion?



#5
Mar2508, 06:48 AM

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. 


#6
Mar2508, 07:26 AM

P: 1,136

[itex]\frac{1}{k(k+1)}=\frac{1}{k}\cdot\frac{1}{(k+1)}[/itex] or [itex]\equiv[/itex]? 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. 


#7
Mar2508, 08:10 AM

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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: [tex]a_{n+1}\,=\,\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}... + \frac{1}{n+1}[/tex] 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? 


#8
Mar2508, 05:02 PM

P: 25

How would i put it into a partial fraction?



#9
Mar2508, 05:59 PM

P: 1,136




#10
Mar2508, 08:00 PM

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? 


#11
Mar2608, 04:19 AM

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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? 


#12
Mar2608, 05:56 AM

P: 25

Alright, i got some more info on the question.
It seems that: [tex]\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}... + \frac{1}{n+1} = 1\frac{1}{n+1}[/tex] But. The number after [tex] \frac{1}{n+1}[/tex] would be [tex] \frac{(n+1)}{(n+1)(n+2)}[/tex] So. [tex]1  \frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)}[/tex] = 1  ?? Would would i simplify the rest of it? 


#13
Mar2608, 06:05 AM

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P: 26,148

ah! … you have a misprint …
it should be [tex]\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}... + \frac{1}{n*(n+1)}\,=\,1\frac{1}{n+1}\,.[/tex] Does that look better? 


#14
Mar2608, 06:06 AM

P: 25

Oh whoops. Yeah thats what i meant



#15
Mar2608, 06:54 AM

P: 25

How would i do the next part?



#16
Mar2608, 07:12 AM

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P: 26,148

[tex]1  \frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)}[/tex]
= [tex] \frac{n+1}{(n+1)}  \frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)}[/tex] = … ? (btw, wouldn't it have been easier to work out if you'd "gone down one", and writen: [tex]1  \frac{1}{n} + \frac{1}{n(n+1)}[/tex] ?) 


#17
Mar2608, 07:31 AM

P: 25

did you just substitute the 1 for a [itex]\frac{n+1}{n+1}[/itex]?
Im really stuck on simplifying the RHS. 


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