( 1)tan(π/4n)>π/4n & gt； 1/4n

The original sequence diverges because ∑( 1/4n) diverges.

(2)n/(n^3+2)= 1/(n^2+2/n)<； 1/n^2

Because ∑ (1/n 2) converges, the original series converges.

(3) 1/[(2n- 1)* 2n]& lt； 1/(2n- 1)^2<； = 1/n^2

Because ∑ (1/n 2) converges, the original series converges.

(4)(cosn)^2/n√n<； = 1/n^(3/2)

Because ∑ [1/n (3/2)] converges, the original series converges.

(5)lim(n->∞)[(n+ 1)/e^(n+ 1)]/(n/e^n)

= lim(n->； ∞)( 1+ 1/n)/e

= 1/e

& lt 1

So the original series converges.

(6)lim(n->∞)[(n+3)/2^(n+ 1)]/[(n+2)/2^n]

= lim(n->； ∞)[(n+3)/2(n+2)]

= lim(n->； ∞)[( 1+3/n)/2( 1+2/n)]

= 1/2

& lt 1

So the original series converges.

(7)lim(n->∞)[5^(n+ 1)/(n+ 1)！ ]/(5^n/n！ )

= lim(n->； ∞)5/(n+ 1)

=0

So the original series converges.

(8)lim(n->∞)[(3^n*n！ )/(n^n)]^( 1/n)

= 3 * lim(n-& gt； ∞)(n！ /n^n)^( 1/n)

= 3 * lim(n-& gt； ∞)[( 1/n)*(2/n)*...*(n/n)]^( 1/n)

= 3 * lim(n-& gt； ∞)e^ln{[( 1/n)*(2/n)*...*(n/n)]^( 1/n)}

= 3 * lim(n-& gt； ∞)e^{( 1/n)*[ln( 1/n)+ln(2/n)+...+ln(n/n)]

=3*e^[∫(0, 1)lnxdx]

=3*e^[(xlnx-x)|(0, 1)]

=3*e^(- 1)

& gt 1

So the original series diverged.