Question

Sum of power series! HELP!

Answer 1

Hint: Read about the Taylor expression for \(\log(x)\), compare your sum with that and then state your results.

Answer 2

isit use this equation?\[\ln x= \sum_{n=1}^{\infty} \frac{ (-1)^{n-1}(x-1)^n }{ n } = (x-1)- \frac{ (x-1)^2 }{ 2 }+\frac{ (x-1)^3 }{ 3 }- \frac{ (x-1)^4 }{ 4 }+...\]

Answer 3

\[\sum_{n=1}^{\infty} -\frac{ (1-x)^n }{ n }= \sum_{n=1}^{\infty} \frac{ (x-1)^n }{ n }\]
i'm stuck i dont know how to incorporate the (-1)^(n-1)

Answer 4

I would do
\[ \int_0^{0.87} \frac{d}{dx} \sum_{n=1}^{\infty} -\frac{(1-x)^n}{n} \ dx\]

Answer 5

swapping the order of the derivative and summation
\[ \int_0^{0.87} \sum_{n=1}^{\infty} -\frac{d}{dx} \frac{(1-x)^n}{n} \ dx\]
\[ \int_0^{0.87} \left(\sum_{n=0}^{\infty} (1-x)^n \right)\ dx\]

Answer 6

the inner summation has a closed form
\[ \frac{ 1 - (1-x)^{\infty}}{1- (1-x)} \]

knowing that 1-x is <0 this simplifies to
\[ \frac{1}{x} \]

Answer 7

and you have
\[ \int_0^{0.87} \frac{1}{x} dx \]

Answer 8

@phi, does this mean, ln0.87-ln0 ?

Answer 9

i still dont get how you shift n=1 to n=0, i would appreciate you explanation on this(:

Answer 10

I messed up the limits on the integral... ln(0) is undefined. Let me tweak my solution, and use an indefinite integral.

Here is what I did
\[ \int \sum_{n=1}^{\infty} -\frac{d}{dx} \frac{(1-x)^n}{n} \ dx \\
\int \sum_{n=1}^{\infty} -\frac{n(1-x)^{n-1}}{n} \ dx\\
\int \sum_{n=1}^{\infty} -(1-x)^{n-1}\ dx
\]

Now let m= n-1, and write the sum as
\[ \int \sum_{m=0}^{\infty} -(1-x)^{m} \ dx\]
replace the summation with its closed-form and integrate
\[ \int \frac{1}{x} \ dx \\ =\ln(x)+ C\]

We know the original sum is zero when x=1, so we know
\[ \sum_{n=1}^{\infty} -\frac{(1-x)^n}{n} \bigg |_{x=1} = 0 = \ln(1)+C \\
0= C\]

The constant of integration C is zero, and we have
\[ \sum_{n=1}^{\infty} -\frac{(1-x)^n}{n} = \ln(x) \]

finally , evaluate ln(0.87)= -0.139

Answer 11

@phi thank you! that was really helpful! :D

Answer 12

btw when you talk about a 'closed-form' how did you get \[\frac{ 1-(1-x)^n }{ 1-(1-x)} \] where did "1" come from? isit from the m= n-1?

Answer 13

\[ \sum_{m=0}^{\infty} -(1-x)^{m} \]

let r= 1-x and move the - sign out front.
\[ - \sum_{m=0}^{\infty}r^m \]
the sum is given by
https://en.wikipedia.org/wiki/Geometric_series#Formula

Answer 14

oooooohhhhh! haha thank you its clear now! thanks for the help! :D

Answer 15

Ouch, I am being sloppy!
There is a minus floating around that should have gone away:
\[ \int \sum_{n=1}^{\infty} -\frac{d}{dx} \frac{(1-x)^n}{n} \ dx \\
\text{ the chain rule means we do d/dx (1-x) = -1 and the -1 goes away}\\
\int \sum_{n=1}^{\infty} \frac{n(1-x)^{n-1}}{n} \ dx\\
\int \sum_{n=1}^{\infty} (1-x)^{n-1}\ dx
\]
Now we can do the summation to get 1/x , and finally integrate dx/x to get ln(x)