CPM Homework Help : CALC Problem 10-136

Determine if the following series converge absolutely, converge conditionally, or diverge. Be sure to indicate which convergence test you used.

$\displaystyle\sum _ { k = 2 } ^ { \infty } ( - 1 ) ^ { k } \frac { k - 1 } { k ^ { 2 } - k }$
$\displaystyle\sum _ { k = 2 } ^ { \infty } \frac { ( - 1 ) ^ { k - 1 } } { \sqrt { k } - 1 }$
$\displaystyle\sum _ { k = 1 } ^ { \infty } \frac { ( - 1 ) ^ { k } ( k + 1 ) ! } { \operatorname { ln } ( k + 1 ) }$
$\displaystyle\sum _ { k = 0 } ^ { \infty } \frac { ( - 1 ) ^ { k } ( k + 1 ) ! } { 2 ^ { 4 k } }$

Step 1 (a):

$\sum_{k=1}^{\infty}\Big|(-1)^k\frac{k-1}{k^2-k}\Big|=\sum_{k=1}^{\infty}\frac{k-1}{k(k-1)}$

Step 2 (a):

$=\sum_{k=1}^{\infty}\frac{1}{k}\text{ which diverges}$

Therefore series does not converge absolutely.

Does this series converge conditionally?

Step (b):

$\lim_{k\to\infty}\frac{1}{\sqrt{k}-1}=0$

Therefore this series converges conditionally by the Alternating Series Test.

Does this series converge absolutely?

Step (c):

$\lim_{k\to\infty}\frac{(k+1)!}{\ln(k+1)}=\infty$

Step (d):

$\lim_{k\to\infty}\frac{(k+1)!}{2^{4k}}=?$