_________________________________________________________________________________
Calculus  II                                                        Quiz #11                                                  May 13, 2003

Name____________________                      R.  Hammack                                              Score ______
_________________________________________________________________________________

Find the interval of convergence of the folloing power series.

(1)    FormBox[RowBox[{Underoverscript[∑ , k = 1, arg3], 5^k/k^2, x^k, Cell[]}], TraditionalForm]

Using the ratio test for absolute convergence:
Underscript[lim , k∞] (| (5^(k + 1) x^(k + 1))/(k + 1)^2 |)/(| (5^kx^k)/k^2 |) = ... ^(k + 1))/(| x |^k) = Underscript[lim , k∞] 5 (k/(k + 1))^2 | x | = 5 | x | = ρ

Thus, for convergence, we must have 5 | x | <1, or   -1/5<x<1/5.

For x = 1/5, the series is  FormBox[RowBox[{Underoverscript[∑ , k = 1, arg3], 5^k/k^2, (1/5)^k, Cell[=], Underoverscript[∑ , k = 1, arg3], 1/k^2}], TraditionalForm]which is a convergent p-series. Thus the series converges for x = 1/5.


For x = -1/5, the series is  FormBox[RowBox[{Underoverscript[∑ , k = 1, arg3], 5^k/k^2, (-1/5)^k, Cell[=], Underoverscript[∑ , k = 1, arg3], (-1)^k/k^2}], TraditionalForm]which is a convergent alternating series. Thus the series converges for x = -1/5.


Thus, the interval of convergence is [-1/5, 1/5].


(2)    Underoverscript[∑ , k = 2, arg3] x^k/ln(k)

Using the ratio test for absolute convergence:
Underscript[lim , k∞] (| x^(k + 1)/ln(k + 1) |)/(| x^k/ln(k) |) = Underscript[li ... 4;] (1/k)/(1/(k + 1)) | x | = Underscript[lim , k∞] (k + 1)/k | x | = | x | = ρ

Thus, for convergence, we must have | x | <1, or   -1<x<1.

For x = 1, the series is  FormBox[RowBox[{Underoverscript[∑ , k = 2, arg3], , 1^k/ln(k), Cell[=], Underoverscript[∑ , k = 1, arg3], 1/ln(k)}], TraditionalForm]which is divergent by comparison with the divergent harmonic series Underoverscript[∑ , k = 2, arg3] 1/k,   as 1/k<1/ln(k).

For x = -1, the series is  Underoverscript[∑ , k = 2, arg3] (-1)^n/ln(k)which is a convergent alternating series. Thus the series converges for x = -1.

Thus, the interval of convergence is [-1, 1).