Find intervals of convergence

I have a problem where the interval of convrsion is (0,2]

However, for the end points plugging the x=0, and x=2 back in is where i am having prolems.

I cannot figure it out what to do after i plug the x in. Can you someone help?

$\displaystyle f(x) =\sum_{n=0}^\infty \frac {(-1)^{n+1}(x-1)^{n+1}} { n+1}$

I will attempt to solve this problem using the ratio

$\displaystyle f(x) =\sum_{n=0}^\infty \frac {(-1)^{n+2}(x-1)^{n+2}} {n+2} . \frac { n+1} {(-1)^{n+1}(x-1)^{n+1}}$

I will attempt to properly cancel out.

$\displaystyle \lim_{n\to\infty} { | (x-1)|} < 1$

then i get for an answer $\displaystyle 0,2$

so i plug x = 2 in to the original equation

$\displaystyle f(x) =\sum_{n=0}^\infty \frac {(-1)^{n+1}(2-1)^{n+1}} { n+1}$

Which is

$\displaystyle f(x) =\sum_{n=0}^\infty \frac {(-1)^{n+1}(1)^{n+1}} { n+1}$

From there i dont know what to do!

Can anyhelp me finish this and do the x = 0 also?

Thank you

Re: Find intervals of convergence

Quote:

Originally Posted by

**icelated** I have a problem where the interval of convrsion is (0,2]

However, for the end points plugging the x=0, and x=2 back in is where i am having prolems.

I cannot figure it out what to do after i plug the x in. Can you someone help?

$\displaystyle f(x) =\sum_{n=0}^\infty \frac {(-1)^{n+1}(x-1)^{n+1}} { n+1}$

I will attempt to solve this problem using the ratio

$\displaystyle f(x) =\sum_{n=0}^\infty \frac {(-1)^{n+2}(x-1)^{n+2}} {n+2} . \frac { n+1} {(-1)^{n+1}(x-1)^{n+1}}$

I will attempt to properly cancel out.

$\displaystyle \lim_{n\to\infty} { | (x-1)|} < 1$

then i get for an answer $\displaystyle 0,2$

so i plug x = 2 in to the original equation

$\displaystyle f(x) =\sum_{n=0}^\infty \frac {(-1)^{n+1}(2-1)^{n+1}} { n+1}$

Which is

$\displaystyle f(x) =\sum_{n=0}^\infty \frac {(-1)^{n+1}(1)^{n+1}} { n+1}$

From there i dont know what to do!

Can anyhelp me finish this and do the x = 0 also?

Thank you

To show the series is divergent when $\displaystyle \displaystyle \begin{align*}x = 0 \end{align*}$, substitute $\displaystyle \displaystyle \begin{align*} x = 0 \end{align*}$ into your series

$\displaystyle \displaystyle \begin{align*} f(x) &= \sum_{n = 0}^{\infty}\frac{(-1)^{n+1}(x-1)^{n+1}}{n + 1} \\ f(0) &= \sum_{n = 0}^{\infty}\frac{(-1)^{n + 1}(0-1)^{n+1}}{n + 1} \\ &= \sum_{n = 0}^{\infty}\frac{(-1)^{n + 1}(-1)^{n + 1}}{n + 1} \\ &= \sum_{n = 0}^{\infty}\frac{\left[(-1)^{n + 1}\right]^2}{n + 1} \\ &= \sum_{n = 0}^{\infty}\frac{1}{n + 1} \end{align*}$

This is the Harmonic Series, which is well known to be (and very easy to show that it is) divergent.

Re: Find intervals of convergence