Math Is Fun Forum

davidi

Member

Registered: 2008-11-05

Posts: 2

Help with tricky infinite series!

Hi all. I wonder if anyone out there can help me find the sum of this infinite series (or at least point me in the right direction):

I know that it converges, because I tested on the Ratio Test:

Thanks in advance

Edit: Fixed my Ratio Test.

Last edited by davidi (2008-11-06 08:26:42)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Help with tricky infinite series!

Now you can use a similar method to write that new summation as some constant (or maybe you already know a formula that will do that )

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Why did the vector cross the road?
It wanted to be normal.

davidi

Member

Registered: 2008-11-05

Posts: 2

Re: Help with tricky infinite series!

Awesome, I think you solved it!
5S looks like a geometric series to me...

Thanks again for the help.

Edit: fixed S = 6/25.

Last edited by davidi (2008-11-06 08:25:03)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Help with tricky infinite series!

Nearly, but that 5 on the denominator gets squared rather than cancelled.
S = 6/25.

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Why did the vector cross the road?
It wanted to be normal.

davidi

Member

Registered: 2008-11-05

Posts: 2

Re: Help with tricky infinite series!

Haha, yes, of course. It's been a long day

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Help with tricky infinite series!

Mathsy’s method is the based on the general method used to evaluate series of the form

(assuming that it converges) where the

form an arithmetic progress and the

form a geometric progression. (In your example,

.)

Suppose the common difference is

and the common ratio is

. Then the series can be written as

If you multiply both sides by

, you get

Subtracting the second from the first yields

If

, the RHS converges to

Hence

In your example,

; hence

.

Last edited by JaneFairfax (2008-11-07 08:16:56)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Help with tricky infinite series!

I'm curious about this now. I think that if you have a sum that looks like:

(where p(x) is a polynomial and |a|<1 to ensure convergence), then the method I posted will write this as something involving:

(where q(x) is a different polynomial whose degree is strictly less than that of p)

Therefore, using this until you end up with a polynomial of degree 0 (a constant), it should be possible to evaluate that top summation for any p and a that fit the restrictions.

But is there a nice formula, or another way to do it without recursion?

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Why did the vector cross the road?
It wanted to be normal.