Determine whether the series converges absolutely,conditionally or diverges

sigma from1to infinity sin((pi/2)n)/ n

To answer this, let's write sin((pi/2)n) in some other (more recognizable) form. sin(pi/2) = 1, sin(pi) = 0, sin(3pi/2) = -1, sin(2pi) = 0, sin(5pi/2) = 1, etc.

So for n = 1, 5, ..., we have sin((pi/2)n) = 1
and for n = 3, 7, ..., we have sin((pi/2)n) = -1
Use this to transform the original series...
Sum_{n=1 to infinity} sin((pi/2)n) / n
= 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + ...
= Sum_{k=1 to infinity} (-1)^(k-1) / (2k-1)
This would have the property that if k is even (i.e. terms with a negative sign), the denominators (in sequence) would be 3, 7, ... (if k=2, the denominator is 3, if k=4, the denominator is 7, and etc.)
Also, if k is odd (i.e. terms with a positive sign), the denominators (in sequence) would be 1, 5, ... (if k=1, the denominator is 1, if k=3, the denominator is 5, etc.)

Sum_{k=1 to infinity} (-1)^(k-1) / (2k-1)
is an alternating sum, the limit of 1/(2k-1) as k-> infinity is zero, and the sequence b_k = 1/(2k-1) is decreasing. So by the alternating series test, it converges (conditionally). Further, if we check for absolute convergence,
Sum_{k=1 to infinity} 1/(2k-1)
is divergent (limit comparison test, p-test, integral test), so this series is not absolutely convergent.

Thus the series is conditionally convergent.

sin(πn/2) = (-1)ⁿ⁺¹ for n ≥ 1. Then we can use the Alternating Series Test. Clearly 1/n -> 0 as n -> ∞, and since
n+1 > n ⇒ 1/n > 1/(n+1)

and so it is monotone decreasing. Therefore the series converges at least conditionally. However
∑ 1/n

is a divergent p-series, as seen by the integral test, so
∑ (-1)ⁿ⁺¹ / n

only converges conditionally.