# Does $\sum_{n=1}^{\infty}(-1)^n\frac{n}{\sqrt{n^3 + 6}}$ converge?

Test the series for convergence or divergence. $$\sum_{n=1}^{\infty}(-1)^n\frac{n}{\sqrt{n^3 + 6}}$$

Because this is an alternating series, I decided to use the alternating series test. This theorem states that if $\lvert a_n \rvert = b_n$ satisfies the conditions:

$b_{n+1} ≤ b_n$ for all $n$

$\lim \limits_{n \to \infty} b_n = 0$

This expression clearly approaches $0$ because $n$ grows asymptotically slower than the denominator of the expression, $\sqrt{n^3 + 6}$, as $n$ approaches $\infty$.

But I think the expression fails to meet the first condition.

While $b_{n+1} ≤ b_n$ is true for all $n>1$, I don't think it is true for $n$. Consider the following values of $b_n$ with values substituted in:

When $n=1$: $$\frac{1}{\sqrt{1^3 + 6}} = \frac{2}{\sqrt{7}} \approx 0.377964473$$ When $n=2$: $$\frac{2}{\sqrt{2^3 + 6}} = \frac{2}{\sqrt{14}} \approx 0.534522483$$ When $n=3$: $$\frac{3}{\sqrt{3^3 + 6}} = \frac{2}{\sqrt{33}} \approx 0.5222329679$$

So, since $0.377964473 < 0.534522483 > 0.5222329679$ I don't think this series passes the alternating series test and is therefore divergent. But I got this question wrong. Apparently, this series converges.

What am I doing wrong? Thanks for your help!

• What happens with the first $N = 3$ or $300$ or $3,000,000$ terms doesn't matter. The sum of the first $N$ terms is always finite. The question is: does there exist an $N$ such that the tests hold for all $n > N$? (Yes!) Apr 14, 2015 at 23:48
• Not only does it converge, it's absolutely convergent (meaning that the absolute value series converges too). Your only problem is that your test values are too early in the sequence. Apr 14, 2015 at 23:48
• This series is not absolutely convergent. Apr 14, 2015 at 23:49
• Okay that makes sense. According to the theorem, I thought it had to hold true for all values of $n$. Really, the theorem should be: $b_{n+1}≤b_n$ for all $n>N$. Apr 14, 2015 at 23:50
• @JamesTaylor Yes! In fact, the series $\sum_{n \geq 1} a_n$ converges iff the series $\sum_{n \geq N} a_n$ is so converges! Jun 12, 2022 at 15:22

Remember exactly what the alternating series test says:

If [some conditions hold] then the series converges.

So, we can never use the alternating series test to conclude that a series diverges; the theorem is silent on the subject of divergence.

The key idea is to find an $N$ such that the absolute value of the terms is monotonically decreasing. That is, if there's some $N$ such that $|a_{n+1}| < |a_n|$ for all $n > N$, then we can use the alternating series test on the sequence

$$\sum_{n = N+1}^\infty a_n,$$

to learn that it converges, and hence the original series must as well (as it's a finite number of terms added to a convergent sequence).

Now your job is to find the $N$ such that the absolute value of terms are monotonically decreasing for $n > N$.