Question

Another series question please:
is the series: (sigma from n=2 to infinity)(-1)^n(4^-n) convergent or divergent, if convergent, what is the sum of the series

Answer 1

\[\sum_{n=2}^{\infty}(-1)^n\frac{ 1 }{ 4^n }\]

Answer 2

I see that n=2, which we usually don't have, we usually have n=0, but I don't know why you're supposed to change that.

Answer 3

\[\sum_{n=0}^{\infty}(-1)^{n+3}\frac{ 1 }{ 4^{n+2} }\]

Answer 4

?

Answer 5

Now Im confused by the exponent "n" appearing outside the bracket and in the denominator

Answer 6

You're making things too hard for yourself (take 2)
\[\huge \sum_{n=2}^{\infty}(-1)^n\frac{ 1 }{ 4^n }=\sum_{n=2}^{\infty}\left(-\frac{ 1 }{ 4 }\right)^n\]

Answer 7

damn, overcomplication indeed

Answer 8

I hope you have it from here :)

Answer 9

why would she tell us to take the n=2 down to zero. I hate when profs show you innecesary pellet

Answer 10

I guess it's not that hard...
\[\huge \sum_{n=2}^{\infty}(-1)^n\frac{ 1 }{ 4^n }=\sum_{n=0}^{\infty}\left(-\frac{ 1 }{ 4 }\right)^{n+2}\]

Answer 11

Tell Professor Wendy I said hi ^_^

Answer 12

uh oh, I just realized I typed in the wrong equation. Its
\[\sum_{n=2}^{\infty}(-1)^{n+1}/4^n\]

Answer 13

As if that's an issue :P
\[\huge \sum_{n=2}^\infty \frac{(-1)^{n+1}}{4^n}=(-1)\sum_{n=2}^\infty \frac{(-1)^{n}}{4^n}\]

Answer 14

ok, i see. youre negating the effect of the plus 1 in the exponent by taking out a neg 1 to the front

Answer 15

So that the numerator and denominator have the same exponent :D

Answer 16

right, thats what I said ;)

Answer 17

So, we're done?

Answer 18

ty!!!!!!

Answer 19

for now. I now thanks to you know how to finish this one

Answer 20

Tell Professor Wendy she's getting too old... :D

Answer 21

will do.

Answer 22

^_^