# Is it possible to maniuplate this infinite sum (series)?

• Mar 18th 2013, 08:44 PM
director
Is it possible to maniuplate this infinite sum (series)?
Hey all,

I have $\sum_{n=0}^{ \infty } a_{n} k^{-n}$ for k=1,2,3... where $a_{n}$ is a term of some infinite sequence { $a_{n}$}

Some manipulation:

$\sum_{n=0}^{ \infty } a_{n} k^{-n} = \sum_{n=0}^{ \infty } \frac {a_{n}} {k^{n}} = \sum_{n=0}^{ \infty } a_{n} \frac {1} {k^{n}} = \sum_{n=0}^{ \infty } a_{n} (\frac {1}{k})^{n}$

Is there anything I can do with this now?

$= \sum_{n=0}^{ \infty } a_{n} \sum_{n=0}^{\infty} (\frac {1}{k})^{n}$

Is it legal to multiply the sums like this?

The second factor looks like a geometric series and I should be able to get the sum of the terms equal to some constant A (if series converges).

$= A \sum_{n=0}^{ \infty } a_{n}$ (for each k)

Does that seem reasonable or did I pull it out of my behind? (Lipssealed)
• Mar 19th 2013, 08:37 PM
chiro
Re: Is it possible to maniuplate this infinite sum (series)?
Hey director.

You can't, in general, do what you have tried to do. Fubini's theorem allows you to exchange the order of summations, but not to do what you tried.

Its the same reason that Integral (fg)dx != Integral (fdx) Integral (gdx) holds for many integrals.

Does a_n have any additional structure or constraints?
• Mar 19th 2013, 08:56 PM
director
Re: Is it possible to maniuplate this infinite sum (series)?
Thanks Chiro!

The two other pieces of info are:

(1) sum(a_n) converges absolutely

(2) $\sum_{n=0}^{ \infty } a_{n} k^{-n} = 0$

I need to prove that all terms of {a_n} must be zero. I think I see why that is but I don't know how to start a formal proof.
I know that the terms of {a_n} must be getting close to 0 since the partial sums converge.

I can think of a sequence with non-zero terms such that (2) is true when k = 1 but false when k = 2.
In general, if such sequence makes (2) true for some k, I would expect it to not satisfy (2) for k+1.
• Mar 19th 2013, 09:36 PM
chiro
Re: Is it possible to maniuplate this infinite sum (series)?
If the sum converges absolutely, then it means that Sum(over n) |a_n| * (1/k)^n = c where c is some finite number.

Since (1/k) is positive, then it means that all terms are greater than or equal to 0.

So the only way you can get zero is if |a_n| = 0 for all n since we know (1/k)^n > 0 for all k which is a natural number.

Formally this is equivalent to showing that the sum is greater than or equal to zero and that you only get c = 0 if all |a_n| = 0.