Proving convergence of the series $\sum_{n=1}^{\infty}\frac {1}{1000n^2 + 25n - 5}$ with comparison test, what can you jump to

I'm just going to make up a series as an example.

$$\sum_{n=1}^{\infty}\frac {1}{1000n^2 + 25n - 5}$$

Not sure if it is a valid series or not, but that's not my question. So if I had to go and prove a series converges by using the comparison test. Say I wanted to compare it to $$\sum_{n=1}^{\infty}\frac {1}{n^2}$$ which I know converges. Would it be too far fetched to compare these two series?

If this was a valid series, then my proof would look like this:

$$\sum_{n=1}^{\infty}\frac {1}{1000n^2 + 25n - 5}$$ $\le$ $$\sum_{n=1}^{\infty}\frac {1}{n^2}$$ Since p > 1, this series in particular converges by p-test, and by comparison test, the original series converges as well.

So yeah, main question, can I jump straight to $$\sum_{n=1}^{\infty}\frac {1}{n^2}$$ from the original series? If not how would I go about doing so?

• Have you heard of the Limit Comparison Test? That is what you should do with your two series. Goes very easy... Commented Mar 30, 2017 at 3:07
• Yes. $1000n^2+25n-5\ge n^2$ for all $n\ge 1$. Commented Mar 30, 2017 at 3:31

Essentially yes, you can just jump to $\sum \frac 1 {n^2}$ since the leading order behavior in the denominator is a constant multiple of $n^2$. To make this rigorous you could do something like this: consider, for $n \ge 1$, we have $25n-5\ge 0$. Thus we have $$\frac{1}{1000n^2+25n-5} \le \frac{1}{1000n^2}.$$ Thus we have $$\sum_{n=1}^\infty \frac{1}{1000n^2+25n-5} \le \sum^{\infty}_{n=1} \frac{1}{1000n^2} = \frac{1}{1000} \sum^\infty_{n=1} \frac{1}{n^2} < \infty.$$