Find the convergate-area to the following series:

∞

SUM (-1)^n * x^n / n(n-1)

n=2

∞

SUM n / 2^n * (x-1)^n

n=1

∞

SUM (-1)^n * 4n * x^n / n ln n

n=2

Struggling a bit with these, so I would highly appreciate ANY help, thanks.

Results 1 to 3 of 3

- Dec 2nd 2007, 08:57 AM #1

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- Dec 2nd 2007, 05:05 PM #2

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∞

SUM (-1)^n * x^n / n(n-1)

n=2

Im just learning this topic so bear with me.

You can solve this by using the ratio test.

Absolute value of A(n+1)/A(n)

The Absoulte value of (-1)^n goes away since its just 1.

so its:

x^n+1 -----n(n-1)

------- * ---------

n+1(n) -----x^n

cross cancel you get

x limit of n -> infinity (n-1)/(n+1) which ='s 1

abs of x < 1 = -1 < x < 1

then plug -1 and 1 in and test if they converge of not.

The 2nd problem you can solve by doing the root test.

∞

SUM n / 2^n * (x-1)^n

n=1

if you root the equation to the n, the power n cancel out

= n/2 * x-1

you can move x-1 outside of the series and have lim n -> infinity n/2 which = infinity only if x is not 1.

If anyone who can double check me would be great. Again, Im not 100% sure, Im still learning this topic.

The third one you can do by using the ratio test again. Try it.

- Dec 3rd 2007, 10:01 AM #3

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## Re: Power Series

Actually, ff4930, it looks like you made an error in applyring the root test to the second series:

$\displaystyle \sqrt[n]{|a_n|}=\sqrt[n]{|\dfrac{n}{2^n}(x-1)^n|}$

$\displaystyle =\dfrac{\sqrt[n]{n}}{2}|x-1|$

not the $\displaystyle \dfrac{n}{2}|x-1|$ that you have, and

$\displaystyle \lim_{n \to \infty}\sqrt[n]{n} = 1$, so that gives |x-1|<2 for the area of convergence. In fact, this problem is done more easily with the ratio test:

$\displaystyle \left | \frac{a_{n+1}}{a_n} \right | = \left | \cfrac{\cfrac{(n+1)(x-1)^{n+1}}{2^{n+1}}}{\cfrac{n(x-1)^n}{2^n}} \right |$

$\displaystyle = \dfrac{(n+1)|x-1|}{2n}$

and $\displaystyle \lim_{n \to \infty}\dfrac{n+1}{2n} = \tfrac{1}{2}$.

Warrick2236, as for the third one, unless you typed it wrong, it looks as if you have an n in the numerator that can cancel with the n in the denominator.

--Kevin C.