s 10/s5=( 1-q^ 10)/( 1-q^5)= 1+q^5=3 1/32

q=- 1/2

LimSn equals? (x→ infinity) =-(1-q n)/(1+1/2) =-2/3.

2. Original formula =lim(n→ infinity) [12+1+2 2+2 ... n 2+n]/n 3.

= lim(n→infinity)[n(2n+ 1)(n+ 1)/6+n( 1+n)/2]/n ^ 3

=(2+0)(0+ 1)/6+0

= 1/3

3. The original formula =lim(n→ infinity) (1+2n) 2n/2 (n 2+ 1).

=(2+0)/( 1+0)

=2

4. Original formula =lim(n→ infinity) [A2n-1]/[A2n+1] (A > 0)

When 0

Lim(n→ infinity) [A 2n- 1]/[A 2n+ 1]

=(0- 1)/(0+ 1)

=- 1

A> in 1

lim(n→infinity)[a 2n- 1]/[a 2n+ 1]= 1。