Question

find the sum of the series where x < 1

Answer 1

\[\sum_{n=0}^{\infty} x^n\]

Answer 2

@TheSmartOne could you help me out please

Answer 3

i know we can use any number greater than 1 but i dont know which one

Answer 4

Hmm, I am not sure.

Maybe @dan815 , @freckles , @mathmate , or @mathstudent55 can help you. :)

Answer 5

ok hopefully they can

Answer 6

hello mathmate

Answer 7

Hello, This is a geometric series.
@El_Arrow

Answer 8

Are you familiar with geometric series, where the common ratio is x?

Answer 9

how did you know that right off the bat?

Answer 10

somewhat familiar we just started covering it today in class

Answer 11

From one term to the next, you multiply by x, so that's the common ratio.

Answer 12

okay so its like a sequence

Answer 13

Yes, it's a geometric sequence, but we're looking for the sum of the terms, so it is a series.

Answer 14

If we sum to term N, then it becomes
\(S=1+x+x^2+x^3+...,.x^N\)
right?

Answer 15

yes

Answer 16

Now do you know how to factorize:
\(1-x^{N+1}\)

Answer 17

does it have to do something with dividing?

Answer 18

Not in the factorizing part.
Actually
\(1-x^{N+1} = (1-x)(1+x+x^2+x^3+...x^N)\)
If you're not sure, or have doubts, multiply it out to verify.

Answer 19

one method is this one

\(\large \color{black}{\begin{align}
y&=\sum_{n=0}^{\infty} x^n \hspace{.33em}\\~\\
y&=x^0+x^1+x^2+x^3+\cdots +\infty \hspace{.33em}\\~\\
y&=1+x^1+x^2+x^3+\cdots +\infty \hspace{.33em}\\~\\
y&=1+x\left(1+x^1+x^2+x^3+\cdots +\infty\right) \hspace{.33em}\\~\\
y&=1+xy \hspace{.33em}\\~\\
y-xy&=1 \hspace{.33em}\\~\\
y(1-x)&=1\hspace{.33em}\\~\\
y&=\dfrac{1}{1-x} \hspace{.33em}\\~\\
\sum_{n=0}^{\infty} x^n&=\dfrac{1}{1-x} \hspace{.33em}\\~\\
\end{align}}\)

Answer 20

so the sequence will continue till it reaches x^n

Answer 21

\(\large \color{black}{\begin{align}
y&=\sum_{n=0}^{\Large{\color{red}{\infty}}} x^n \hspace{.33em}\\~\\
\end{align}}\)
the upper limit is \(\infty\) not \(n\) so it will go on forever

Answer 22

Actually, if we divide both sides by (1-x), we get
\((1-x^{N+1})/(1-x) = (1+x+x^2+x^3+...x^N)\)
And the sum you are looking for can be found by putting N-> infinity.
Since x<1, x^(N+1)=0, and you will get the result.

The reason I take this approach is because sometimes you are required to sum to term N only, in which case you already for the formula for it.

Answer 23

*already have the formula...

Answer 24

oh okay i see now

Answer 25

what if the n=1 would it still be the same formula

Answer 26

Yes, exactly.
Say, x=1/2
N=1
we have
S=1+x=1+1/2=3/2
or
\(S=(1-x^{N+1})/(1-x)\)
\(=(1-(1/2)^2)/(1-1/2)\)
\(=(1-1/4)/(1-1/2)\)
\(=3/4*2/1\)
\(=3/2\)
as before.