Simplifying a geometric series with an infinity summation bound

[atrus_ovis]

Homework Statement

I am solving some convolutions, and i have come to these solutions.
a)$$\sum$$2^k, summing from -$$\infty$$ to -1

b)$$\sum$$2^k, summing from -$$\infty$$ to n , where n <=-1

Homework Equations

the geometric series summation formula, from 0 to N

$$\sum$$a^k = 1-a^N+1 / 1-a , summing from 0 to N

The Attempt at a Solution

Is there a direct formula to simplify the sums I've come to?I mean a formula for any summation bounds, not just for "from zero to N" format.Moreover, can i plug in infinity to the formula mentioned above?

I understand that i can break up the sum and/or flip the sum bounds any way that i like, correct?

 

[Mentallic]

$$\sum_{k=0}^{\infty}a^k=\frac{1}{1-a}$$ iff |a|<1

So in other words, if -1<a<1, then its infinite summation will be finite.

In the case where the summation is from $$-\infty$$ to 0, we have to consider for what values of a it will be finite. If we check $$\lim_{k\to-\infty}a^k$$ it is obvious that this is equivalent to $$\lim_{k\to \infty}a^{-k}=\lim_{k\to\infty}(\frac{1}{a})^k$$ so in this case, we need |1/a|=1/|a|<1, thus |a|>1.

Can you see any way of how this sheds light on your problem?

 

[atrus_ovis]

But the summation is not from -$$\infty$$ to 0, but from
-$$\infty$$ to n, and from -$$\infty$$ to -1.

Does the following move me forward?

$$\sum$$2 ^k from -$$\infty$$ to n
= $$\sum$$2 ^-k from -n to $$\infty$$
= $$\sum$$(1/2) ^k from -n to $$\infty$$

similarly

$$\sum$$2 ^k from -$$\infty$$ to -1
=$$\sum$$(1/2) ^k from 1 to $$\infty$$

Should i break them down like this:
$$\sum$$(1/2) ^k from -n to $$\infty$$
=$$\sum$$(1/2) ^k from -n to 0
+ $$\sum$$(1/2) ^k from 0 to $$\infty$$ and
(edited)
$$\sum$$(1/2) ^k from 1 to $$\infty$$
= $$\sum$$(1/2) ^k from 0 to $$\infty$$
- $$\sum$$(1/2) ^k from 0 to 1(also, how can i correctly write the sum notation with tex?)

 

[Mentallic]

Yes exactly To find $$\sum_{k=-\infty}^{n}$$ you simply split it up into $$\sum_{k=-\infty}^{0}+\sum_{k=1}^{n}$$ and then switch the infinite sum as you've done. Now you should finish with explaining in mathematical terms that the only way the sum is going to be finite is if the infinite sum has a ratio of size less than 1.

You can click on my latex to see how it's done.

 

[atrus_ovis]

Okay, i REALLY can't write in notation.
So:
$$<br /> \sum_{k=-\infty}^{-1}<br />$$ 2^k = $$<br /> \sum_{k=1}^{\infty}<br />$$ (1/2) ^k=$$<br /> \sum_{k=0}^{\infty}<br />$$ - (1\2)⁰ (because k=0 isn't included in the sum, but we have to start from 0 to apply the formula)

= lim_{m->$$\infty$$} { $$<br /> \sum_{k=0}^{m}<br />$$(1/2)^k }-1

=lim_{m->$$\infty$$} {2 * [1 - (1/2)^m+1] -1} = 1 , since limit of 2^-m, with m approaching infinity is 0)Is the methodology/result correct?

edit: and the sum from -infinity to n, results after two sum break-ups, to 2ⁿ⁺¹

 

[Mentallic]

You should put the entire equation in latex, such as

$$\sum_{k=-\infty}^{-1}2^k=\sum_{k=1}^{\infty}\left(\frac{1}{2}\right)^k$$

Fractions can be written as \frac{numerator}{denominator} and you can put brackets in just as they are () but they're usually too small, so I just throw in a \left( and \right) which makes the brackets suitably bigger.

For limits, $$\lim_{m\to\infty}$$ just click on my latex to see how to write them.

(and by the way, the previewing doesn't work for latex so you will need to submit your reply and then edit the post if necessary)

The formula doesn't need to be started from k=0. You can use $$\frac{a(1-r^{n+1})}{1-r}$$ for a geometric summation that starts at the value a and has a ratio of r. If you start at k=0, your a will be 1.

But either way, it still works so I guess you can use whatever you're most comfortable with.

Yes what you have seems good.

 

[atrus_ovis]

and by the way, the previewing doesn't work for latex

Ah!
That drove me batgarbage insane!

Thanks,for both the math and Latex help.

 

[Mentallic]

Ah!
That drove me batgarbage insane!

Yeah I also had a big wtf moment when it first screwed up. I type something in latex and check to see if it's right, and it instead gives me something totally different and I was drunk at the time so that didn't help either. I sat there redoing my latex thinking how the hell this is happening hehehe

Thanks,for both the math and Latex help.

No worries!