Question

(Calculus) Need help with a series question!

Answer 1

One second, I'm typing up the problem..

Answer 2

\[\sum_{1}^{\infty} (-1)^(n+1)*1/2^n\]

Answer 3

I expressed that as : (1/2)-(1/2^2)+(1/2^3)-....

Answer 4

your question is to find whether it is convergent or div
and find the sum?

Answer 5

Then I factored out a 1/2. So it became, 1/2(1-(1/2)+(1/2^2)-(1/2^3)+...

Answer 6

\[1/2\sum_{1}^{\infty} r^n\]

Answer 7

$$ \Large { \sum_{n=1}^{\infty} (-1)^{n+1}\left(\frac 1 2 \right)^n }$$

Answer 8

So, I'm am trying to figure out what my "r" term would be..

Answer 9

is that \[\sum_{1}^{\infty}\frac{(-1)^{n+1}}{2^n}\]

Answer 10

In the "new" summation.

Answer 11

you can absorb the -1

Answer 12

$$ \Large {
\sum_{n=1}^{\infty} (-1)^{n+1}\left(\frac 1 2 \right)^n
\\=\sum_{n=1}^{\infty} (-1)\cdot (-1)^{n}\left(\frac 1 2 \right)^n
\\=\sum_{n=1}^{\infty} (-1)\cdot\left(\frac {-1}{2} \right)^n

}$$

Answer 13

\[1/2\sum_{1}^{\infty}r^n = 1/2(1-(1/2)+(1/2^2)-(1/2^3)+...)\]

Answer 14

That's what I have, but I can't figure out with the "r^n" would be..

Answer 15

r = -1/2

Answer 16

Wow, that makes so much more sense now! :)

Answer 17

Well, would the r still be the same in the summation that I mentioned earlier (for the \[1/2\sum_{1}^{\infty} r^n?\]

Answer 18

I'm basically trying to express the original summation in the form a very simply geometric series.

Answer 19

$$ \Large {
\sum_{n=1}^{\infty} (-1)^{n+1}\left(\frac 1 2 \right)^n
\\=\sum_{n=1}^{\infty} (-1)\cdot (-1)^{n}\left(\frac 1 2 \right)^n
\\=\sum_{n=1}^{\infty} (-1) \cdot \left( (-1) \cdot \frac 1 2 \right)^n

\\=\sum_{n=1}^{\infty} (-1)\cdot\left(\frac {-1}{2} \right)^n
\\=-\sum_{n=1}^{\infty}\cdot\left(\frac {-1}{2} \right)^n
= - \frac{\frac{-1}{2}}{1 - \frac{-1}{2}}
= \frac{\frac{+1}{2}}{1 + \frac{+1}{2}}
\\= \frac{\frac{1}{2}}{\frac{3}{2}}
=\frac13
}$$

Answer 20

what was your question again, sorry working on Latex

Answer 21

Hah, One second let me retype it :)

Answer 22

\[\sum_{1}^{\infty} (-1)^(n+1) *1/2^n = (1/2)-(1/2^2)+(1/2^3)-...\]

Answer 23

From that, \[1/2(1-(1/2)+(1/2^2)-(1/2^3)+...\]

Answer 24

yes that works, that is a geometric series

Answer 25

and from that, \[1/2\sum_{1}^{\infty} r^n \] what would be r in this case?

Answer 26

r would be -1/2

Answer 27

even with the 1/2 outside of the summation?

Answer 28

note that if you divide consecutive terms , you get -1/2

Answer 29

$$ \Large r = \frac {a_{n+1}}{a_n}$$

Answer 30

yes its ok to have 1/2 outside the sum

Answer 31

Okay, so. \[1/2\sum_{1}^{\infty} (-1/2)^n \]

Answer 32

is equal to \[1/2(1-(1/2)+(1/2^2)-(1/2^3)+...)\]

Answer 33

its fine to bring it inside the sum, however. it will change your 'first' term
$$ \Large {

\sum_{n=0}^{\infty} a\cdot r^n= \frac{a}{1-r}
} $$

Answer 34

'a' is your first term, in the series expansion

Answer 35

important caveat, this sum is valid only if |r | < 1

Answer 36

Would you mind showing me the series expansion for it?

Answer 37

left side does not equal to right side

Answer 38

$$\large
1/2(1-(1/2)+(1/2^2)-(1/2^3)+...) \color{red}\neq 1/2\sum_{n=1}^{\infty} (-1/2)^n
\\ ~\\~\\
\large
1/2\sum_{n=1}^{\infty} (-1/2)^n = 1/2[(-1/2)^1+(-1/2)^2+(-1/2)^3+...)
$$

Answer 39

if you start at n=0, you can fix that

Answer 40

Thanks a bunch!