Question

I don't understand this, please help. Determine if the series is convergent or divergent:
If convergent, find the sum othe series.
(sigma)from n=0 to infinity of (-1)^n*3/4^n

Answer 1

This series
\[\huge \sum_{n=0}^{\infty}(-1)^n\left(\frac{3}{4}\right)^n\]?

Answer 2

converges like anything

Answer 3

Except something divergent, @satellite73 :)

Answer 4

true that

Answer 5

okay smarties,thanks. But how do I show this? What "test" do I use, I just don't understand it :(

Answer 6

a) \((\frac{3}{4})^n\to 0\) as \(n\to \infty\) and the series alternates, so that is enough

Answer 7

Who needs tests? Without the (-1)^n, it's just a geometric series with a common ratio less than 1

So it's absolutely convergent :D

Answer 8

but it has the (-1)^n, so why isnt that significant?

Answer 9

b)\( \sum(\frac{3}{4})^n=\frac{1}{1-\frac{3}{4}}=4\)

Answer 10

well peter pan, "Professor Wendy" requires tests

Answer 11

I mean, the series is absolutely convergent without the (-1)^n
so with the (-1)^n, it would still be convergent :)

Answer 12

it makes it smaller so it converges for sure

Answer 13

That's annoying of her :)
Use the Alternating Series test, see if the sequence converges to zero :)

Answer 14

or compare with
\[\sum(\frac{3}{4})^n=\frac{1}{1-\frac{3}{4}}=4\]

Answer 15

since it is smaller than 4, it converges

Answer 16

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

Answer 17

Can only compare positive series with the comparison test, @satellite73
Naughty naughty...

Answer 18

For instance, this series can also be compared with your series @satellite73
but it's obviously divergent
\[\huge \sum_{n=1}^{\infty}-\frac1n\]

Answer 19

If you have a series
\[\huge \sum_{n=k}^{\infty}(-1)^na_n\]

Answer 20

I understand the p-series. But why would I ignore the (-1)^n part?

Answer 21

Where a_n is a positive series, if the limit of a_n as n becomes really really big is zero, the series
\[\huge \sum_{n=k}^{\infty}(-1)^na_n\]

Answer 22

is convergent :)

Answer 23

But if you don't like that, there's another way. :)

Answer 24

Thats super. TY. Then, what is the sum? Is that the limit?

Answer 25

What is the other way?

Answer 26

No. I'm afraid getting the sum isn't THAT simple :)

Answer 27

crap, ok how?

Answer 28

The other way is the root test :)
Given a series
\[\huge \sum_{n=k}^{\infty}a_n\]if
\[\huge \lim_{n\rightarrow \infty}\sqrt[n]{|a_n|}<1\]

Answer 29

Then the series is absolutely convergent :)

Answer 30

yeah, no on the root test here

Answer 31

Okay, then stick to the Alternating series test :)

Answer 32

what is the sum however? I am not sure

Answer 33

why can't i compare?

Answer 34

Actually, our series may be written as
\[\huge \sum_{n=0}^{\infty}\left(-\frac{3}{4}\right)^n\]

Answer 35

@satellite73
Were you using the direct Comparison test?

Answer 36

yes

Answer 37

Well, you may only compare two series
\[\huge \sum_{n=k}^{\infty}a_n\]and
\[\huge \sum_{n=k}^{\infty}b_n\]
If both a_n and b_n are positive series.

Answer 38

oh i see, the negative terms. in any case the terms go to zero, not minus infinity or something

Answer 39

I suppose, the absolute values of the terms of the series are exactly the same as the terms of a known convergent series,
comparison test, you get equality, you get absolute convergence,

Boom, convergence ^.^

Answer 40

The sum!!!

Answer 41

add two series might work

Answer 42

Need the sum!!!

Answer 43

I told you, @jotopia34
The series is just like
\[\huge \sum_{n=0}^{\infty}\left(-\frac{3}{4}\right)^n\]

Answer 44

oh better idea

Answer 45

And it's a geometric series with first term 1
and common ratio -3/4

Answer 46

so I do a/1-r

Answer 47

And while you're thinking about that, let's play with infinite geometric series :)
The sum can be written
\[\Large S=a+ar+ar^2+ar^3+ar^4+...\]\[\Large S-a=ar+ar^2+ar^3+ar^4+ar^5+...\]\[\Large S-a=r(a+ar+ar^2+ar^3+ar^4+...\]\[\Large S-a=rS\]\[\Large S-rS=a\]\[\Large S(1-r)=a\]\[\Large S=\frac{a}{1-r}\]

Answer 48

Oh, sorry, what were we talking about? ^.^

Answer 49

lol reinventing the wheel again

Answer 50

Because I can >:)

Answer 51

lol, so I take it the answer is yes to my sum solving proposal

Answer 52

hey, @satellite73
What do you mean by "again" ? >.>

Answer 53

And yes, @jotopia34 ^.^