Help with MaTh

1. You want the geometric series to converge, so -1 < r < 1.

2. 1 - 2 + 3 - 4 + ... + 2005 - 2006 + 2007 = (1 - 2) + (3 - 4) + ... + (2005 - 2006) + 2007 = 1002.

3. The product will telescope, and you are left with n/(n - 1), so you want n/(n - 1) < 1.98.  The largest n that satisfies this is n = 201.

1)
Common ratio = $\dfrac{6r}{2}$ = 3r

-1 < (Common ratio) < 1

-1 < 3r < 1

$\dfrac{-1}{3} < r < \dfrac{1}{3}$

2)

Consider the pairwise sum of terms.

$\quad 1-2+3-4+\cdots+2005-2006+2007\\ =(1-2)+(3-4)+\cdots+(2005-2006)+2007\\ =-1+(-1)+\cdots+(-1)+2007\\ \text{There are in total }\dfrac{2006}{2}=1003\text{ pairs that sums to }-1.\\ = -1(1003) + 2007\\ = 1004$