Question

Find the sum of the following infinite geometric series, if it exists.

Answer 1

\[\frac{ 1 }{ 2 } + (-\frac{ 1 }{ 4 }) + \frac{ 1 }{ 8 } + (-\frac{ 1 }{ 16 })...\]

Answer 2

@KeithAfasCalcLover

Answer 3

So we have a infinite sum:
\[\sum^\infty_{n=0}{\frac{(-1)^{n}}{2^{n+1}}}\]

Answer 4

okay

Answer 5

what am i supposed to do with that

Answer 6

Wait lol
Ehh im not sure...
Hold up I've tried couple things...

Answer 7

AHA! GOT IT\[\sum^{\infty}_{n=0}{\frac{(-1)^n}{2^{n+1}}}=\sum^{\infty}_{n=0}{\frac{(-1)^n}{2^{n}\times2}}=\frac{1}{2}\sum^{\infty}_{n=0}{\left(-\frac{1}{2}\right)^n}\]
We know that:
\[\sum^\infty_{k=0}{ar^k}=\frac{a}{1-r}\]
So we can simplify:
\[\frac{1}{2}\sum^\infty_{n=0}{(1)\left(-\frac{1}{2}\right)^n}=\frac{1}{2}\times\frac{1}{1-(-\frac{1}{2})}=\frac{1}{2}\times\frac{1}{1+\frac{1}{2}}=\frac{1}{2}\times\frac{1}{3/2}=\frac{1}{2}\times\frac{2}{3}=\frac{1}{3}\]

Answer 8

And that is the final answer. I can verify it
http://www.wolframalpha.com/input/?i=infinite+sum+of&f1=(-1)%5En%2F(2%5E(n%2B1))&f=Sum.sumfunction_(-1)%5En%2F(2%5E(n%2B1))&f2=0&f=Sum.sumlowerlimit%5Cu005f0&f3=infinity&f=Sum.sumupperlimit%5Cu005finfinity&a=*FVarOpt.1-_**-.***Sum.sumvariable---.*--

Answer 9

Is that what you were looking for? :)