# prove that $\sum_{3}^{\infty}\frac{n+1}{n^3-3n+9}$ converges

the question is: prove that $$\sum_{n=3}^{\infty}\frac{n+1}{n^3-3n+9}$$ converges

I tried the ratio test but it equals 1 and i tried direct comparison but $$\frac{1}{x^2}$$ is too small and i couldn't find any other risenable guess the integral test will work but it is not a "nice" solution (computing that integral by hand will be unpractical for me i tried splitting it in to n/... and 1/... and using partial fractions but it didn't lead anywhere) over all im at a loss as to how to approach this kind of problem any suggestions as to my method( at this point its just guessing and trying convergence tests) will be appreciated

• Use the limit comparison test with $\displaystyle\sum_{n=3}^\infty \dfrac{1}{n^2}.$ – John Thompson May 4 at 21:24

Let $$n \ge 3$$.

$$\dfrac{n+1}{n^-3n+9} \lt \dfrac{n+1}{n^3-3n} \lt$$

$$\dfrac{2n}{n(n^2-3)} = \dfrac{2}{n^2-3} \lt$$

$$\dfrac{2}{n^2-(1/2)n^2} = \dfrac{4}{n^2}.$$

Comparison test ($$\sum \dfrac{1}{n^2}$$ converges).

Use the comparaison test: it converges because $$\sum_{n=1}^\infty\frac1{n^2}$$ converges and$$\lim_{n\to\infty}\frac{\frac{n+1}{n^3-3n+9}}{\frac1{n^2}}=1.$$

Hint: $$n+1\lt 2(n-1)$$ and $$n^3-3n+9\gt (n-1)^3$$ for large enough $$n$$ - and your "too small" is no longer a problem. Notice how an apparently complicated situation is simplified by using quite crude estimates - the constant factor of $$2$$ is not a problem.

Note that you only need estimates for large $$n$$ (finite initial parts of sums don't affect eventual convergence).