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 *ab**solutely 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.**