Method of Differences for Finding Summation with Fractions - Homework Help

[thomas49th]

Homework Statement

I'm given the identity:
$$\sum{\frac{r^{3}-r+1}{r(r+1)} \equiv r-1 + \frac{1}{r} - \frac{1}{r+1}$$

EDIT: from r=1 to n on the sum sign

Homework Equations

I know I'm ment to use the method of differences and I know that

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

The Attempt at a Solution

I presume you need to find out how you can relate the 3 fractions to the summation of r?

Thanks :)

 

[tiny-tim]

Homework Statement

I'm given the identity:
$$\sum_{r=1}^n{\frac{r^{3}-r+1}{r(r+1)} \equiv r-1 + \frac{1}{2} - \frac{1}{r+1}$$

Hi thomas49th!

Hint:

i] r³ - r = … ?

ii] with a view to using differences, 1/r(r+1) = … - … ?

 

[thomas49th]

sorry to of wasted your time but the orginal question was

$$\sum{\frac{r^{3}-r+1}{r(r+1)} \equiv r-1 + \frac{1}{r} - \frac{1}{r+1}$$

there is no 1/2

sorry!

Errr can I combine it to this:
$$\sum (r) - 1 + \sum(\frac{1}{r} - \frac{1}{r+1})$$

does that help at all?

$$\sum(\frac{1}{r} - \frac{1}{r+1})$$
simplifies to $$\sum(\frac{1}{r^{2} + r})$$

Thanks! :)

 

[tiny-tim]

$$\sum(\frac{1}{r} - \frac{1}{r+1})$$
simplifies to $$\sum(\frac{1}{r^{2} + r})$$

So ∑ 1/(r² + r) = … ?

sorry to of wasted your time but the orginal question was

$$\sum{\frac{r^{3}-r+1}{r(r+1)} \equiv r-1 + \frac{1}{r} - \frac{1}{r+1}$$

there is no 1/2

sorry!

erm … I could see it was wrong! … in fact, I still think it's wrong!

Errr can I combine it to this:
$$\sum (r) - 1 + \sum(\frac{1}{r} - \frac{1}{r+1})$$

does that help at all?

Not following you.

The method of difference involves subtracting the answer for r = n from the answer for r = n-1.

 

[thomas49th]

It's question number 4

I could do part a) that's easy just part part b) is giving me trouble.

At first I tried $$\sum (r) - 1 + \frac{1}{\sum (r)} - \frac{1}{\sum (r) + 1}$$

but that's wrong?

Thanks :)

 

[tiny-tim]

At first I tried $$\sum (r) - 1 + \frac{1}{\sum (r)} - \frac{1}{\sum (r) + 1}$$

The ∑n part is right.

The 1 should be ∑1.

The rest is just ∑(1/r - 1/r+1) …

that's already in differences, isn't it? …

so its sum is … ?

 

[thomas49th]

is the sum:
$$\frac{n(n+1)}{2} - n + \frac{2}{n(n+1)} - \frac{2}{n(n+1) + 2n}$$

is that right?

is the sum of 1 = n?

 

[tiny-tim]

is the sum:
$$\frac{n(n+1)}{2} - n + \frac{2}{n(n+1)} - \frac{2}{n(n+1) + 2n}$$

is that right?

is the sum of 1 = n?

The first two terms are right.

As I said before: The rest is just ∑(1/r - 1/r+1)

Look … that's (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + …

can you see what's happening?

this is why it's called the method of differences, and why it works!

 

[thomas49th]

I know the trick when things cancel out.
So are you substituting values 1 to n into 1/2 - 1/r+1?

how was I ment to spot that you substitute values into there and not the first part?

Thnanks :)

 

[tiny-tim]

So are you substituting values 1 to n into 1/r - 1/r+1?

Exactly!

how was I ment to spot that you substitute values into there and not the first part?

Because the first part was easy … it was just ∑(r + 1) …

but the second part was ∑ 1/r(r+1), which obviously needs a trick!

 

[thomas49th]

Ahh yes I got it

$$\frac{n^3 + n}{2(n+1)}$$

Thank you ever so much :)