Series, find Divergence or Convergence

[CitizenInsane]

Homework Statement

Find the Divergence or Convergence of the series

$\sum$$^{∞}_{n=1}$$\frac{2n^2+3n}{\sqrt{5+n^5}}$

Homework Equations

Ratio Test, Comparison Test, Limit Comparison Test, Integral test etc.

The Attempt at a Solution

This question was on my final exam and the only question of which I couldn't actually figure out. I tried the ratio test multiple times and ended getting L=1 which is no info multiple times. I figured the comparison test was the right approach but had no idea what series to compare it to.

 

[DryRun]

You could compare it to: $$\frac{2n^2}{\sqrt{n^5}}$$Or you could try factorizing it by taking out $n^2$ from the numerator and denominator.

 

[CitizenInsane]

You could compare it to: $$\frac{2n^2}{\sqrt{n^5}}$$Or you could try factorizing it by taking out $n^2$ from the numerator and denominator.

Can you expand on your suggestions? I am utterly clueless with respect to this problem.

 

[DryRun]

Taking out $n^2$ from the numerator and denominator:$$\sum^{\infty}_{n=1}\frac{2+\frac{3}{n}}{\sqrt{5/n^4 + n}}$$Well, the nth-term test is not useful here, as the limit is 0.

 

[CitizenInsane]

Taking out $n^2$ from the numerator and denominator:$$\sum^{\infty}_{n=1}\frac{2+\frac{3}{n}}{\sqrt{5/n^4 + n}}$$Well, the nth-term test is not useful here, as the limit is 0.

Ye I was about to say that lol.

 

[DryRun]

Ye I was about to say that lol.

You should always try with the simpler tests first.

Using the comparison test: $$u_n=\frac{2n^2+3n}{\sqrt{5+n^5}}$$$$v_n=\frac{2n^2}{\sqrt{n^5}}=\frac{2}{n^{1/2}}$$Now, use the p-series test and the series $v_n$ diverges.
Since $v_n \le u_n$ and $\sum^{\infty}_{n=1}v_n$ diverges, therefore the original series diverges.

 

[CitizenInsane]

You should always try with the simpler tests first.

Using the comparison test: $$u_n=\frac{2n^2+3n}{\sqrt{5+n^5}}$$$$v_n=\frac{2n^2}{\sqrt{n^5}}=\frac{2}{n^{1/2}}$$Now, use the p-series test and the series $v_n$ diverges.
Since $v_n \le u_n$ and $\sum^{\infty}_{n=1}v_n$ diverges, therefore the original series diverges.

Thanks.