Limit Comparison Test on $\sum_{n=1}^{\infty}\frac{2n^2-1}{3n^5+2n+1}$

Question

I have the series:

$\sum_{n=1}^{\infty}\frac{2n^2-1}{3n^5+2n+1}$

I compared it with: $\sum_{n=1}^{\infty}\frac{n^2}{n^5}$

The limit is $\frac{2n^5-n^3}{3n^5+2n+1}$ as $n$ approaches infinity.

The limit works out to $\frac23$. I don't understand how this shows convergence? It is less than 1, so why is the p-series, the second series I have, converging?

Answer 1

You mixed up comparison test and limit comparison test.

Limit comparison test:

Consider two series of non negative reals $\sum u_n$ and $\sum v_n$ if

$lim_{n\to \infty} \frac{u_n}{v_n}=l \space (0<l< \infty)$ then both the series converge or diverge simultaneously.

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$u_n=\frac{2n^2-1}{3n^5+2n+1}$

Let $v_n=\frac{1}{n^3}$

Then $\lim_{n\to \infty}\frac{\frac{2n^2-1}{3n^5+2n+1}}{\frac{1}{n^3}}=\frac{2}{3}$

Since $\sum \frac{1}{n^3}$ converge ( $\sum\frac{1}{n^p}$ converges for $p>1$ ) , by limit comparison test $\sum\frac{2n^2-1}{3n^5+2n+1}$ also converge.

Note: Convergence of $p$- series.

Answer 2

The limit is correct. However, since the p-series will diverge if p=1,(since a p-series diverges if 0<p<=1), then the finite, positive limit shows the p-series will diverge. If the p-series diverges, then by the Limit Comparison Test, the other series must diverge as well.