Question

Calculate the serie

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

Solution

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

\[\sum_{n=0}^{\infty}((\frac{1}{2})^{n})-\sum_{n=0}^{\infty}(\frac{-3}{4})^{n}.(\frac{-3}{4}^{2}) = \]
\[\sum_{n=0}^{\infty} \frac{1}{2}^{n} -\sum_{n=0}^{\infty} \frac{-3}{4}^{n}.\sum_{n=0}^{\infty}\frac{9}{16} = \frac{9}{16}
\]
can you check it out if the solution correct is?

Answer 1

\[\sum_{n=0}^{\infty}\frac{1}{2^n}=2\] for the first term

Answer 2

hmm why ? i thought its \[0\]

Answer 3

\[\sum_{n=0}^{\infty}(\frac{3}{4})^n=\frac{1}{1+\frac{3}{4}}=\frac{7}{4}\] but you have to subtract off the first two terms

Answer 4

you thought what was zero?

Answer 5

i thought \[ \sum_{n=0}^{\infty} (2^{-n})\] goes to \[0\]

Answer 6

ok hold on
\[\lim_{n\to \infty}\frac{1}{2^n}=0\] for sure

Answer 7

otherwise there is no way that the sum would be finite

Answer 8

but you are adding
\[1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.
..\] and that is clearly not zero, it is 2

Answer 9

aah ok

Answer 10

using the formula for the sum of the geometric series
\[\sum x^n=\frac{1}{1-x}\] with \(x=\frac[1}{2}\)

Answer 11

\(x=\frac{1}{2}\)

Answer 12

that is also how you add up the second series, but you have to be careful, because you are missing the first two terms
you are starting at \(n=2\) not \(n=0\) so you sum them using the formula above, subtract off the first two terms, and then was ever you get you subtract from the number \(2\)

Answer 13

*whatever you get

Answer 14

sorry something i thought i understand but i didnt
why \[1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.
.. = 2 \] ? e.g. why \[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.
.. = 1 \] ?

Answer 15

you want the reason behind the formula, or the formula?

Answer 16

is there any formula for \[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.
.. = 1 \] is ?

Answer 17

formula for
\[a+ar+ar^2+ar^2+...\] is
\[\frac{a}{1-r}\]

Answer 18

lots of ways to see this. for one thing algebra will show you that
\[(1-x)(1+x+x^2+x^3+...+x^n)=1-x^{n+1}\] so
\[1+x+x^2+x^3+...+x^n=\frac{1-x^{n+1}}{1-x}\] and if \(|x|<1\) we have as \(x\to \infty\) \(x^n\to 0\) leaving
\[\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}\]]

Answer 19

another way is to let
\[S=a+ar+ar^2+ar^3+...\] then put
\[rS=ar+ar^2+ar^3+...\] so that
\[S-rS=a\] and so
\[S=\frac{a}{1-r}\]

Answer 20

hmm ok i try to understand it....

Answer 21

the problem you posted presupposes you know how to sum a geometric series.

Answer 22

because to solve the problem you have to sum two of them, then subtract

Answer 23

hmm ok i look your advices and try to solve it again.. thank you very much

Answer 24

look over what i wrote because the method is there
first sum is 2
second sum i also computed above
the only thing left to do is to subtract off the first two terms from second sum, because you are starting at \(n=2\)
then whatever you get , subtract that from 2

Answer 25

that is
\[\sum_{n=2}^{\infty}\left(-\frac{3}{4}\right)^n=\sum_{n=0}^{\infty}\left(-\frac{3}{4}\right)^n-(1-\frac{3}{4})\]

Answer 26

ok i hope i can solve it if not it would be much appreciate if you can solve it for me :)

Answer 27

i have already more or less
\[\sum_{n=2}^{\infty}\left(-\frac{3}{4}\right)^n=\sum_{n=0}^{\infty}\left(-\frac{3}{4}\right)^n-(1-\frac{3}{4})\]
\[=\frac{1}{1+\frac{3}{4}}-\frac{1}{4}=\frac{1}{\frac{7}{4}}-\frac{1}{4}=\frac{4}{7}-\frac{1}{4}=\frac{9}{28}\]
subtract from 2

Answer 28

ok is it end solution satellite or i do need subtract it from 2 ?

Answer 29

aah ok i get it

Answer 30

yeah you have to subtract it form two

Answer 31

i need \[2-\frac{9}{28}\]

Answer 32

thank you very much