quiz10.nb

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Calculus  II                                                         Quiz #10                                                 May 9, 2003

Name____________________                     R.  Hammack                                             Score ______
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Decide if the following series converge or diverge. Use any applicable test.

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

Notice that $FormBox[RowBox[{RowBox[{sin^2(k), (1/2)^k, Cell[]}], ≤, RowBox[{(1/2)^k, Cell[]}]}], TraditionalForm]$ , and    $FormBox[RowBox[{Underoverscript[∑ , k = 1, arg3], (1/2)^k, Cell[]}], TraditionalForm]$ is a convergent geometric series.
Thus, the given series converges by the comparison test.

(2)

Using the ratio test, Underscript[lim , k∞] 3^(k + 1)/(k + 1) !/3^k/k ! = Underscript[lim , k∞] 3^(k + 1)/(k + 1) ! k !/3^k = Underscript[lim , k∞] 3/(k + 1) = 0<1 , so the series converges.

(3)

Notice that $FormBox[RowBox[{RowBox[{RowBox[{5/(e^k + 1), Cell[]}], ≤, 5/e^k}], =, RowBox[{5, (1/e)^k, Cell[]}]}], TraditionalForm]$ , and    $FormBox[RowBox[{Underoverscript[∑ , k = 1, arg3], 5, (1/e)^k, Cell[], Cell[]}], TraditionalForm]$ is a convergent geometric series.
Thus, the given series converges by the comparison test.

(4)

Using the ratio test for absolute convergence,

Underscript[lim , k∞] (| (-1)^(k + 1) (k + 1)^3/3^(k + 1) |)/(| (-1)^kk^3/3^k |) ... 4;] (k + 1)^3/3^(k + 1) 3^k/k^3 = Underscript[lim , k∞] 1/3 ((k + 1)/k)^3 = 1/3<1

Thus, the series converges absolutely, so it converges.