# Convergeny Again - Direct Comparison

It's me again. Somebody please tell me if its considered as spam what i do here for asking a question so often :)

$$\sum_{n\geq0} \frac{n+3}{7n^2-2n+1}$$

I assume that this series diverges as it seems to be something like the harmonic series.

Considering I take the sequence from my series as b_n, I would like to find a sequence a_n so that a_n <= b_n <= 0 is true.

So if the series a_n diverges, for example $$\sum_{n\geq1}{\frac{1}{n}}$$ i proved that b_n also diverges.

So I have

$$\frac{1}{n} \leq \frac{n+3}{7n^2-2n+1}$$ which is true but only as n >= 2.

I think that isn't enough. Which approach would you guys do?

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why shouldn't it be enough? The series $-1 + \sum (1/n)$ is still divergent. Looseley Speaking: Neglecting a finite number of terms can not turn a divergent series into a convergent one. Didn't see this, why would you want $b_n<0$? –  Quickbeam2k1 Sep 10 '13 at 17:36

In general, if you have another series $\sum_n b_n$ that diverges, and if the following limit exists $\lim_{n \to \infty} a_n / b_n = L \neq 0$, then $\sum_n a_n$ also diverges. You can use $b_n = 1/n$ and you will get a limit of $1/7$ (e.g. you can use L'Hopital's rule to evaluate the limit), so your series diverges.
You have to do it with $1/7n$ instead of $1/n$.
$$\frac{1}{7n}\le \frac{n}{7n^2}\le\frac{n+3}{7n^2-n}\le\frac{n+3}{7n^2-n-(n-1)}=\frac{n+3}{7n^2-2n+1}$$