Question

Find the sum of the geometric series: see comment

Answer 1

Find the sum of the geometric series
\[\sum_{n=1}^{\infty} \frac{ (-3)^{n-1} }{ 4^{n} }\]

I am unsure where to start.

Answer 2

The geometric series is defined as

\[ \sum_{n=0}^{\infty} ar^n \] and will converge so long as |r|<1. notice that n=0, so if we shift to get n=1, we would have to -1 from the power to obtain:
\[ \sum_{n=1}^{\infty} ar^{n-1} \]

The geometric series converges to:

\[ \sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{(1-r)}\]
so long as |r|<1

Now we just need to get it into our a*r form and pick off what we need, so:

\[ \sum_{n=1}^{\infty} \frac{(-3)^{n-1} }{4^n}\] notice that the -3 is risen to the correct power, but the 4 is not, we can rewrite as:

\[ \sum_{n=1}^{\infty} \frac{(-3)^{n-1} }{4^{n-1}*4}\] due to power of exponents, so then, lets rearrange and obtain:

\[ \sum_{n=1}^{\infty} \frac{1}{4}\left(\frac{-3}{4} \right)^{n-1} \]
We can see now that a = 1/4 and r = -3/4. Not that |(-3/4)| is less than 1, so it will converge, then we plug in our and r

\[ \sum_{n=1}^{\infty} \frac{1}{4}\left(\frac{-3}{4} \right)^{n-1} = \frac{\frac{1}{4}}{1+\frac{3}{4}} \]

Then simplify

Answer 3

Excellent explanation. Thank you a ton. :)

Answer 4

No problem, just remember with these problems, manipulation helps.