Let t=x-π, then a (n+ 1) = ∫ (n π, (n+ 1) π)-Sint/(t+π) dt = ∫ (n π, (n+1)

In addition, an=∫(nπ, (n+ 1)π) sinx/xdx.

When n is odd, 0

|a(n+ 1)|=a(n+ 1)

=∫(nπ，(n+ 1)π) -sinx/(x+π)dx

& lt=∫(nπ，(n+ 1)π) -sinx/xdx

=-Ann

=|an|

When n is an even number, 0

|a(n+ 1)|=-a(n+ 1)

=∫(nπ，(n+ 1)π) sinx/(x+π)dx

& lt=∫(nπ，(n+ 1)π) sinx/xdx

= Ann

=|an|

To sum up, | a (n+ 1) | < =|an|

(2)an=∫(nπ，(n+ 1)π) sinx/xdx

According to the integral mean value theorem, there exists k∈(nπ, (n+ 1)π), so: an = π * sink/k.

Because n π

Because lim (n->; ∞)- 1/n = lim(n-& gt； ∞) 1/n=0

According to the limit, lim (n->; ∞)an=0

9 、( 1)[an+a(n+2)]/n =( 1/n)*∫(0，π/4) [(tanx)^n+(tanx)^(n+2)]dx

=( 1/n)*∞(0，π/4)(tanx)^n*[ 1+(tanx)^2]dx

=( 1/n)*∞(0，π/4) (tanx)^n*(secx)^2dx

=( 1/n)*∞(0，π/4) (tanx)^n*d(tanx)

=[ 1/n(n+ 1)]*(tanx)^(n+ 1)|(0,π/4)

= 1/n(n+ 1)

= 1/n- 1/(n+ 1)

Therefore, the original series =1-1/2+1/3+...+1/(n+1)+ ...

= lim(n->； ∞)[ 1- 1/(n+ 1)]

= 1

(2) From the conclusion of the problem (1), we can know that an+a(n+2)= 1/(n+ 1).

Because the sequence {an} is a positive sequence, so

For any λ > 0 with /n λ

According to the properties of P series, ∑ 1/n (1+λ) converges.

Therefore, through comparison and discrimination, ∑ an/n λ converges.