Question

Find the infinite sum

Answer 1

Geometric series?

Answer 2

\[\large \sum_{0}^{\infty} \frac{2^{2n}}{5^n}\]

Answer 3

Just haven't done sums or series in awhile...need a refresher lol

Answer 4

would it be n = 0 at the bottom ?

Answer 5

yeah, I just skipped the n = part..since n is the only variable..but yes

Answer 6

first you test to see if the series converges by finding the ratio \[\frac{a_{n+1}} { a_n} = r\], if r is less than 1, it converges

Answer 7

And for that I get 4/5 so it does

Answer 8

yep, so it does have a limit

..im looking for the infinite serires equation online... having trouble finding it...

Answer 9

The

\[\large \frac{a}{1 - r}\] one?

Answer 10

Omg...that's it -_-

Answer 11

mmm... i think so,
what do you get for the soulution

Answer 12

converges to 5

Answer 13

oh nice! haha ^_^

Answer 14

I swear when I'm doing it on paper I make so many mistakes and yet when I come on here I work it out -_- lol

Thanks @DemolisionWolf

Answer 15

Alright how about another one...

Answer 16

\[\large \sum_{0}^{\infty} \frac{2 + (-3)^n}{4^n}\]
Ratio test gives r = -3/4

plugging in n = 0 I get a = 3

infinite series ...

3 12
-------------- = --------
1 + 3/4 7

But this is wrong...where am I going off?

Answer 17

1 sec

Answer 18

Take your time lol just review :)

Answer 19

\[\frac {2+-3^0}{4^0} = \frac {2+-1}{1} = 1\]

\[\frac {2+-3^0}{4^0} = \frac {2+-1}{1} = 1\

Answer 20

oh wow okay that was a mistake lol...okay well then that gets me to...

1 4
--------- = -------
1 + 3/4 7

Hmm...still dont believe that's correct...did I make an error doing the ratio test?

Answer 21

yeah I think I did...hang on...

Answer 22

mmmm hahaha
i'm rusty on this too haha....

so when
n=0, a_0 = 1
n=1, a_1 = -1/4
n=2, a_2 = -7/16

a_2 = r * a_1
-7/16 = r * (-1/4)
r = 7/4

so r >1, does not converge thus there is no upper limit to the infinate series

Answer 23

That's so odd, my paper says it converges ...hmm... I get that if r >1 it diverges...ehh probably just an error on the paper

Answer 24

i'm loggin off, I got my own homework to do :/

hope it helps!

Answer 25

I 'm pretty sure in order for a series to converge, r must be < 1

Answer 26

abs(r) < 1 that is