I am trying to use the comparison test to determine whether the following infinite series converges.

$$\sum_{n=1}^\infty \frac{1}{\sqrt{n^3+2n-1}}$$

$$\frac{1}{\sqrt{n^3}} > \frac{1}{\sqrt{n^3+2n-1}} $$

Is there a way to show that $1/\sqrt{n^3}$ converges? I used Wolfram|Alpha and it told me it does.

  • 2
    $\begingroup$ Compare it with the integral, for example. $\endgroup$ – Daniel Fischer Oct 29 '13 at 0:06
  • $\begingroup$ Thanks! I don't know why I didn't use the integral test. It does converge. $\endgroup$ – Quaxton Hale Oct 29 '13 at 0:17
  • $\begingroup$ Compare $n^2$ and $n^{3/2}$. $\endgroup$ – wannadeleteacct Oct 29 '13 at 0:57

Yes. The $p$-series theorem says $$\sum_{n=1}^\infty {1\over n^p} < \infty$$ iff $p > 1$.


The power of the denominator is above 1, so it converges. If my memory serves me correctly, it is only necessary to look at the highest-power term (in this case $n^3$). We know that $\sqrt{n^3} = n^{3/2}$, which grows faster than n... Remember, n is the boundary - a higher order denominator will converge, while a lower order denominator will diverge.


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