Proof Question: Using Mathematical Induction

Lococard

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

Gib Z

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.

Lococard

Anyone got any ideas with this problem?

I cant find any notes regarding a similar question.

HallsofIvy

What was wrong with Gib Z's suggestion?

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.

Schrodinger's Dog

Putting it in partial fractions gives.

[latex]\frac{1}{k(k+1)}=\frac{1}{k}\cdot\frac{1}{(k+1)}[/latex] 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.

tiny-tim

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:

[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:
then:

You see how that makes the problem easier?

Lococard

How would i put it into a partial fraction?

Schrodinger's Dog

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.

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?

tiny-tim

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?

Lococard

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?

tiny-tim

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?

Lococard

Oh whoops. Yeah thats what i meant

Lococard

How would i do the next part?

tiny-tim

[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] ?)

Lococard

did you just substitute the 1 for a [itex]\frac{n+1}{n+1}[/itex]?


Im really stuck on simplifying the RHS.