CPM Homework Help : APCALC Problem 10-96

Examine the following series. Use one of the tests you have learned so far to determine if the series converges or diverges. State the tests you used.

$\displaystyle\sum _ { n = 1 } ^ { \infty } ( \frac { 1 } { 2 n } - \frac { 1 } { 3 n } )$
Hint (a):
$\frac{1}{2n}-\frac{1}{3n}=\frac{1}{6n}$

$\displaystyle\sum _ { n = 1 } ^ { \infty } \frac { n ! } { ( 2 n ) ! }$
Step 1(b):
$\lim\limits_{n\to\infty}\frac{(n+1)!/(2n+2)!}{n!/(2n)!}$
Step 2(b):
$=\lim\limits_{n\to\infty}\Bigg(\frac{(n+1)!}{n!}\cdot\frac{(2n)!}{(2n+2)!}\Bigg)$
Step 3(b):
$=\lim\limits_{n\to\infty}\frac{n+1}{(2n+2)(2n+1)}=?$

$\displaystyle\sum _ { n = 1 } ^ { \infty } n e ^ { - n ^ { 2 } }$
Hint (c):
$ne^{-n^2}=\frac{n}{e^{n^2}}$
Step (c):
$\lim\limits_{n\to\infty}\frac{(n+1)/e^{(n+1)^2}}{n/e^{n^2}}$