test for convergence or divergence:
inf(sigma)n=1 [(5n-3)/(n^2-2n+5)]

Question

anonymous · Answer

It converges.  You can replace all of that with (for convergence tests only)$$\sum_{n=1}^{\infty}(1/(n^2)$$ which converges according to the p-test.

Convergent :D

watchmath · Answer

It is divergent. Use the limit comparison test! compare to $b_n=\frac{1}{n}$.
The limit of $a_n/b_n$ is 5 but $\sigma b_n$ is divergent. So the original is divergent as well.

anonymous · Answer

Why would you compare it to 1/n?

watchmath · Answer

Because the degree difference between the top and the bottom is 1. So roughly the series behave like the harmonic series.

anonymous · Answer

Good answer.  Wasn't thinking :D