Verified answer

Assuming that the sum starts at n = 1, break the sum up into

sum(n=1 to infinity)(n/n^2) - sum(n=1 to infinity)(1/n^2) =

sum(n=1 to infinity)(1/n) - sum(n=1 to infinity)(1/n^2).

Now the second of these sums converges to ((pi)^2 / 6), but the first

of the sums is the harmonic series, which is known to diverge.

Edit: A more formal approach would be to use the limit comparison test with

the harmonic series sum(1/n), where we find that

lim(n->infinity)[((n-1)/n^2) / (1/n)] =

lim(n->infinity)((n-1)/n) = lim(n->infinity)(1 - (1/n)) = 1.

Since this value is finite the limit comparison test tells us that the two

series either both converge or diverge. Since the harmonic series diverges,

we can thus conclude that the given series also diverges.

for n>1 it is strictly non-negative. Hence if it converges, it converges absolutely and can be rearranged and/or the sum can be broken up.