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Calculus  II                                           Quiz #8                     April 29, 2005

Name_________________            R.  Hammack                  Score ______
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Decide if the following series converge or diverge. In the case of convergence, say whether the series converges conditionally or absolutely.

(1) Underoverscript[∑ , k = 0, arg3](k^2 - 4)/(k^6 + k + 1)
For k > 1, the series has positive terms.
Further, (k^2 - 4)/(k^6 + k + 1)<k^2/(k^6 + k + 1)<k^2/k^6=1/k^4
Therefore, the series converges by comparison with the convergent p-series Underoverscript[∑ , k = 0, arg3]1/k^4
Since it converges and its terms are all positive, then it also converges absolutely.

(2)   Underoverscript[∑ , k = 1, arg3]2^k/k !
Using the ratio test Underscript[lim , n∞]2^(k + 1)/(k + 1) !/2^k/k !=Underscript[lim , n∞]2^(k + 1)/(k + 1) !k !/2^k=Underscript[lim , n∞]2^(k + 1)/2^kk !/((k + 1) k !)=Underscript[lim , n∞]2/(k + 1)= 0
Therefore the series converges. Since the terms are positive, it converges absolutely


(3)    Underoverscript[∑ , k = 1, arg3](-1)^k/(k + 1)
This is an alternating series, with a_1>a_2>a_3>... and Underscript[lim , n∞]a_k=Underscript[lim , n∞]1/(k + 1)= 0.
Therefore it converges by the alternating series test.
However,   Underoverscript[∑ , k = 1, arg3]|(-1)^k/(k + 1)| = Underoverscript[∑ , k = 1, arg3]1/(k + 1)=1/2+1/3+1/4+... is the (divergent) harmonic series (minus the first term).
Therefore the original series converges conditionally.


(4)    5/6-6/7+7/8-8/9+9/10-10/11+ ...
Note that Underscript[lim , n∞]u_k does not exist, for odd terms approach 1 and even terms approach -1.
Therefore the series diverges by the divergence test.