Proof Question: Using Mathematical Induction

[Lococard]

Homework Statement

Prove that, for all integers $n =>1$

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

Homework Equations

I am a little stuck on this question. :|

The Attempt at a Solution

 

[Gib Z]

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.

 

[Lococard]

Anyone got any ideas with this problem?

I can't 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 I am a pain.

 

[Schrodinger's Dog]

We haven't been taught that.

Im a little unsure what he means as well.

Could you possibly give more help?

Sorry if I am 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.

 

[tiny-tim]

… give it a name … !

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?

 

[Lococard]

How would i put it into a partial fraction?

 

[Schrodinger's Dog]

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.

 

[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]

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?

 

[Lococard]

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?

 

[tiny-tim]

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?

 

[Lococard]

Oh whoops. Yeah that's what i meant

 

[Lococard]

How would i do the next part?

 

[tiny-tim]

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

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

= … ?

(btw, wouldn't it have been easier to work out if you'd "gone down one", and writen:
$$1 - \frac{1}{n} + \frac{1}{n(n+1)}$$ ?)

 

[Lococard]

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

 

[tiny-tim]

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 … ?

 

[Lococard]

How did you do that step?

Now I am really lost.The lecturer suggested i get the equation to = $1 - \frac{1}{n+2}$

 

[tiny-tim]

ah!

You don't like polynomial fractions, do you?

They're just like ordinary fractions …

If you had 1 - 15/17, you'd say that's 17/17 - 15/17, = 2/17.

Similarly, if you had 1 - 15/(n+17), you'd say that's (n+17)/(n+17) - 15/(n+17), = (n+17-15)/(n+17) = (n+2)/(n+17).

You got to practise this, and become happy with it!

So … 1 - 1/(n+1) = … ?

 

[Lococard]

Ah.

1 - $\frac{1}{N+1} = \frac{n+1}{n+1} - \frac{1}{n+1} = \frac{n}{n+1}$How did you get:

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

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

 

[tiny-tim]

1 - $\frac{1}{N+1} = \frac{n+1}{n+1} - \frac{1}{n+1} = \frac{n}{n+1}$

Yes!

How did you get:

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

to get to:

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

I didn't - I was just doing the first two terms.

Now, what is

$$\frac{n}{(n+1)} + \frac{1}{(n+1)(n+2)}$$ ?

 

[Lococard]

That was the bit i was stuck on.

 

[tiny-tim]

ok, well you do the same thing:

n/(n+1) = n(n+2)/(n+1)(n+2).

And then … ?

 

[Lococard]

how do you transfer $$\frac{1}{(n+1)(n+2)}$$ from the LHS to the RHS?

 

[tiny-tim]

I don't follow

From the LHS to the RHS of what?

 

[Lococard]

$$\frac{n}{(n+1)} + \frac{1}{(n+1)(n+2)}$$ = 0

n/(n+1) = n(n+2)/(n+1)(n+2).Did you swap $$\frac{1}{(n+1)(n+2)}$$ to the other side of the equals sign?

How did you simplify did you get n(n+2)/(n+1)(n+2). Is it all on the LHS of the equals sign, or did you swap some of it to the right side?

 

[tiny-tim]

Sorry, you've completely lost me:

$$\frac{n}{(n+1)} + \frac{1}{(n+1)(n+2)}$$ = 0

Where did the "= 0" come from?

There's no 0 in this.

Did you swap $$\frac{1}{(n+1)(n+2)}$$ to the other side of the equals sign?

How did you simplify did you get n(n+2)/(n+1)(n+2). Is it all on the LHS of the equals sign, or did you swap some of it to the right side?

I just did: $$\frac{n}{(n+1)}\,=\,\frac{n}{(n+1)}\,\frac{(n+2)}{(n+2)}\,=\,\frac{n(n+2)}{(n+1)(n+2)}\,.$$

That was so that I could then add it to the $$\frac{1}{(n+1)(n+2)}$$ , giving … ?

 

[Lococard2]

$\frac{n(n+2)}{(n+1)(n+2)}$is that right?

Would i then expand:

$\frac{n^2+2n}{n^2+3n+2}$

then after cancelling:

$\frac{1}{(n+2)}$

 

[tiny-tim]

… stick to the plan …

$\frac{n(n+2)}{(n+1)(n+2)}$

is that right?

Yes!

Would i then expand:

$\frac{n^2+2n}{n^2+3n+2}$

Noooo … never expand the bottom !

then after cancelling:

$\frac{1}{(n+2)}$

Cancelling what?

(And how did you get that?)

The whole plan was to add 1/(n+1)(n+2) to it first … then you can try cancelling.

 

[Lococard2]

So, what i need to do is:

$\frac{n(n+2)}{(n+1)(n+2)} + \frac{1}{(n+1(n+2)}$

Doesnt that just equal:

$\frac{n(n+2)}{(n+1)(n+2)}$?

 

[tiny-tim]

… attack of the clones …

So, what i need to do is:

$\frac{n(n+2)}{(n+1)(n+2)} + \frac{1}{(n+1(n+2)}$

Doesnt that just equal:

$\frac{n(n+2)}{(n+1)(n+2)}$?

Hey … what happened to the original Lococard?

I liked him!

erm … just read what you've written … ​

 

[Lococard]

haha, Sorry i lost my password and this connection doesn't allow gmail.com to load.

But I am back now.

$\frac{n(n+2)+1}{(n+1)(n+2)}$

And from then i simplfy it?

 

[tiny-tim]

… long live lococard … !

haha, Sorry i lost my password and this connection doesn't allow gmail.com to load.

But I am back now.

I can tell the difference!

$\frac{n(n+2)+1}{(n+1)(n+2)}$

And from then i simplfy it?

Yes!

 

[Lococard]

So that is the same as:

$\frac{(n+1)(n+1)}{(n+1)(n+2)}?$

yes? no?

Now I am not sure what to do :S