Geometric series question

[MarkSA]

Hello.

I have a question like this:

Summation from n=1 to infinity of: (-3)^(n-1)/(4^n)
Find whether it converges or diverges, and find the sum if it converges.

The example shows steps of:
Summation (etc) of: [(-3)^(n-1)]/[4.4^(n-1)]
Summation (etc) of: 1/4 * (-3/4)^(n-1)
In a/1-r, a is then 1/4 and r is -3/4. and then it's solvable.

But how are they manipulating the denominator to get 4.4^(n-1)? Why is it legal? I'm having trouble seeing how they do it for some reason. I'm also confused as to why they can pull a 1/4 out in the second step. I would have thought it would still be raised to the exponent n-1 like the -3/4 is?

Thanks

 

[galactus]

\(\displaystyle \sum_{n=1}^{\infty}\frac{(-3)^{n-1}}{4^{n}}\)

Rewrite:

\(\displaystyle \sum_{n=1}^{\infty}\frac{(-3)^{n}(\frac{-1}{3})}{4^{n}}\)

\(\displaystyle \frac{-1}{3}\sum_{n=1}^{\infty}(\frac{-3}{4})^{n}\)

Now, can you finish?.

 

[MarkSA]

Sorry I'm still having trouble seeing how to get from (-3)^(n-1) to (-3)^(n)(-1/3). By plugging in the numbers I see they are equal. But how is it derived? I don't think i've ever manipulated exponents like this unless i'm just missing something obvious.

I see now how they pulled the fraction outside the summation from your steps... im unsure about the first part though.

edit: Actually I think I see it now... Did you rewrite (-3)^(n-1) as (-3)^n/(-3)? Then simply pulled the -1/3 out? Wow that's tricky.

 

[galactus]

Yep, that's it.