# Absolute or conditional convergence of alternated series.

I need to study the convergence (absolute or conditional) of this alternated series :

$\sum\limits_{n=1}^{\infty}(-1)^n \left(\frac{n}{7n^2+3}\right)$

Here is what I did so far :

$\lim\limits_{n \to \infty}\left(\frac{n}{7n^2+3}\right) = \lim\limits_{n \to \infty}\left(\frac{1}{7n+\frac{3}{n}}\right) = 0$

So it would be absolute convergence.

However, it seems way too easy to be the real solution.. Is that correct ? Else what do I need to do ?

• No, it's not, I'm afraid. Commented Jun 9, 2014 at 23:59
• @DavidMitra Ok so with what I already did, I proved that it does converge then I would evaluate with p-series test for $1/n$ and it would not converge since !p>1 which would mean that the series conditionally converge ? Commented Jun 10, 2014 at 0:12
• What you already did above doesn't show directly that it converges. Show that $(n/(7n^2+3))$ is decreasing and has limit $0$. Then the Alternating Series Test will show the series converges. For absolute convergence, compare with $\sum{1\over 14n}$. This series diverges (it's a divergent $p$-series) and so $\sum{n\over 7n^2+3}$ diverges. By sure to show the hypotheses for the Comparison Test are satisfied. Commented Jun 10, 2014 at 0:15
• @DavidMitra The derivative of the function is $\frac{3-7x^2}{(7x^2+3)}$ which is less thant 0 for all x >= 1 then $(n/7n^2+3)$ is decreasing and limits tend to 0 which mean that the series converge. Then I just need to prove that 1/14n diverge using p-series (p is not greater than 1) which mean that $n/7n^2+3$ also diverge. Did I understand right ? Commented Jun 10, 2014 at 0:26
• Yes; that looks good. (You want sums, in your second sentence, of course.) Commented Jun 10, 2014 at 0:47

Rule 1 for series: if $a_n$ does not tend to $0$ then $\sum a_n$ diverges.
Rule 2 for series: the converse of rule 1 is not true. In other words, if $a_n\to0$ then $\sum a_n$ may or may not converge.
For absolute convergence you need to tell whether or not $$\sum_{n=1}^\infty \frac{n}{7n^2+3}$$ converges. Hint: we have $$\frac{n}{7n^2+3}\ge\frac{n}{7n^2+3n^2}=\frac{1}{10}\frac{1}{n}\ .$$ You should know whether $\sum\frac{1}{n}$ converges or diverges, and how to use that fact to solve this problem.
For $$\sum_{n=1}^\infty(-1)^n \frac{n}{7n^2+3}$$ you should use the Alternating series test (Leibniz' test). There are various ways to check the necessary details, but you might start by using differentiation to show that $$f(x)=\frac{x}{7x^2+3}$$ is decreasing for $x\ge1$.