calculus II

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The radius of convergence of a power series is the radius of the largest disk with center ` a` within which the series converges absolutely for a series indexed by `n ` that is a function of a variable `x `.

If the series is written in the form

`f(x)...

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The radius of convergence of a power series is the radius of the largest disk with center ` a` within which the series converges absolutely for a series indexed by `n ` that is a function of a variable `x `.

If the series is written in the form

`f(x) = sum_(n=0)^infty c_n(x-a)^n `

then, according to the root test for convergence, the series converges absolutely inside the open disk whose radius is the radius of convergence if `C<1 ` where

`C = lim _(n -> infty) "sup" (root(n)(|c_n|) |x-a|) `

and where the radius of convergence ` r` is such that

`r = 1/("lim"quad"sup"_(n-> infty) root(n)(|c_n|)) `

Alternatively, according to the ratio test of convergence, the series converges if

`|x-a| <r `

where the radius of convergence `r ` is given by

`r = lim_(n->infty) |c_n/c_(n+1)| `

Here we have `f(x) = sum_(n=1)^infty (-1)^((n+1)) (x-(-2))^n/(n2^n) `   (note that the term for n=0 is excluded as dividing by zero gives an undefined result)

Thus we have that `a=-2 `  and that

`c_n = (-1)^((n+1))/(n2^n) `    

Now,  `r = lim_(n->infty) |c_n/c_(n+1)| = lim_(n->infty) |((-1)^(n+1)(n+1)2^(n+1))/((-1)^(n+2)n2^n)| = lim_(n->infty) |-(2(n+1))/n| ` `= lim_(n-> infty) (2(n+1))/n = 2 ` .

Since the series converges if  `|x-a| <r ` then the ` `interval of convergence must contain the interval

`a - r < x < a + r `  ie here the interval  `-2 - 2 < x < -2 + 2 ` or simply  `-4 < x < 0 ` .

To completely identify the interval of convergence we also check whether it converges at the edges of this interval, ie `x=-4 `  and `x=0 ` . Plug the values in:

For `x=-4 `  we check whether  `f(-4) = sum_(n=1)^infty (-1)^((n+1))(-2)^n/(n2^n) ` converges. We can rearrange this to give

`f(-4) = sum_(n=1)^infty(-1)^((2n+1))1/n `   . Since the series is always negative due to the consistently odd power term involving -1, `x=-4 ` isn't in the interval of convergence.

However, in the case `x=0 ` , we check whether `f(0) = sum_(n=1)^infty (-1)^((n+1))2^n/(n2^n) = sum_(n=1)^infty(-1)^((n+1))/n `   converges. By the alternating series test this does converge. Hence `x=0 ` is in the interval of convergence.

If we are interested in the interval for which the series is absolutely convergent, ie the sum of the absolute value of the terms converges, then these two endpoints `x = -4 `  and  `x=0 `   aren't in the interval of convergence since the terms are always positive.

In terms of conditional convergence, when `x=0 `  `f(x) `  is convergent but not absolutely convergent. The series is said to be conditionally convergent  when `x=0 ` .

Answer

a) radius of convergence r=2

b) interval of convergence -4 < x <= 0 or `x in (-4,0] `

c) interval of absolute convergence -4 < x < 0 or `x in (-4,0) `

d) the series is conditionally convergent where it is convergent but not absolutely convergent. This occurs only when x=0.

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