**Parentheses Matter!**

- completely outside the summation: 1 + ∑
_{n=1}^{∞}((-1)(^{n}*n*+1)^{2}/*n*^{5}) ; - inside the summation, as a separate term: ∑
_{n=1}^{∞}(1 + (-1)(^{n}*n*+1)^{2}/*n*^{5}) ; or - inside the summation, as part of the denominator: ∑
_{n=1}^{∞}((-1)(^{n}*n*+1)^{2}/(*n*^{5}+ 1)) ;

*n*is strictly positive, the summand has the form (-1)

^{n}

*b*

_{n}_{ }where the b

_{n}≥ 0. Thus, we could show that this series converges using the

*Alternating Series Test*, which merely requires that we show that:

- lim
_{n}_{→}_{∞}*b*= 0; and_{n } - {
*b*} is a decreasing sequence._{n}

*Absolutely Convergent*. Thus, to determine this we must examine the series consisting of the absolute values of the original terms, which in this case is just ∑

_{n}

_{=1}

^{∞}

*b*If this series converges, then the original series is converges absolutely.

_{n}*Comparison Test.*

- for
*n*>2*,*(*n*+1)^{2}<*n*^{3}; and - (
*n*+1) >^{5}*n*^{5}

*n*+ 1)

^{2}/(

*n*+ 1) <

^{5}*n*

^{3}/

*n*1/

^{5}=*n*

^{2}

^{∞}

_{n=1}_{ }(1/

*n*

^{2}) converges.

^{2}/6 (pi squared divided by 6).

*Integral Test*, or more directly, by the

*p-series Test,*

^{∞}

_{n=k}(1/

*n*) for

^{p}*k > 0*( = 1 in our case) and

*p > 1*(= 2 in our case).