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

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1. anonymous

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

2. watchmath

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.

3. anonymous

Why would you compare it to 1/n?

4. watchmath

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

5. anonymous

Good answer. Wasn't thinking :D

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