Math Is Fun Forum

simplyjasper

Member

Registered: 2009-11-15

Posts: 24

Need help in finding this infinite series

How should I approach finding the sum of this infinite series

Last edited by simplyjasper (2009-11-15 04:14:11)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Need help in finding this infinite series

Hi simplyjasper;

That is an indefinite summation. For those you must use the summation by parts formula.

To use that formula you must understand both the difference calculus and the summation calculus. Notice that it is very similar to the integration by parts formula.

If it were like this, a definite summation, you would have a few more options.

This one is much easier, is this what you intended?

Last edited by bobbym (2009-11-15 06:02:18)

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

simplyjasper

Member

Registered: 2009-11-15

Posts: 24

Re: Need help in finding this infinite series

Hi simplyjasper;

That is an indefinite summation. For those you must use the summation by parts formula.

To use that formula you must understand both the difference calculus and the summation calculus. Notice that it is very similar to the integration by parts formula.

If it were like this, a definite summation, you would have a few more options.

This one is much easier, is this what you intended?

Yes that is what i'm intending to find. Had no idea how to do that as i'm still new to posting these formula in this forum... How do you approach this question and manage to find it's 1?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Need help in finding this infinite series

Hi simplyjasper;

Please be careful to provide the intervals of summation in the future because:

Now to the definite summation problem:

There is a simple way to do a definite sum like this and you should always look for it.

First notice that the summand, can be represented as a difference.

which means:

Now manually expand the 2 right hand sums:

Now look at all those fractions on the RHS of A) and B). If you subtract B from A it is intuitively obvious that all the fractions will cancel leaving only the 1. So:

Last edited by bobbym (2009-11-15 22:03:11)

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

simplyjasper

Member

Registered: 2009-11-15

Posts: 24

Re: Need help in finding this infinite series

Appreciate your reply...
Did you use 'telescoping'? As after a great deal of time and random trial and error,  I did manage to get this partial fractions...

But I doubt I have such a luxury of time during my exam...

I was just wondering is there any technique in helping you find a the relationship faster or even another way of solving the sum of this series

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Need help in finding this infinite series

Hi simplyjasper;

Yes, that is my own form of telescoping

Here is some info on telescoping.

http://en.wikipedia.org/wiki/Telescoping_series

But I doubt I have such a luxury of time during my exam...

I am probably not the best person to ask on this because most of these type of sums are like second nature to me. Here is some advice:

1) You can always use the method of partial fractions on those rational forms. This is slow for everyone.
2) Test problems are made to be solved. They will always have a nice neat solution. Look for it.
3) Practice on a few more and post them and I will try to increase your speed.
4) Yes, there are other ways but they are slow also and require more math.

Last edited by bobbym (2009-11-16 09:52:02)

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.