Question 1164143
<pre>
0, 3/5, 4/15, 5/25, 6/35, 7/45, 8/55, 9/65

The general term is 

{{{a[n] = expr(1/2)((n+1)/(5(2n-3))+abs((n+1)/(5(2n-3)))))(n+1)/5(2n-3)/2}}} for n=1,2,...

"Half the sum of the 'would-be term' and its absolute value" causes
only the first term to be 0, yet keeps all other terms as following this:

{{{(n+1)/(5(2n-3))}}}

Edwin</pre>