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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!

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    $\begingroup$ 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!) $\endgroup$
    – Simon S
    Apr 14, 2015 at 23:48
  • $\begingroup$ 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. $\endgroup$
    – Brian Tung
    Apr 14, 2015 at 23:48
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    $\begingroup$ This series is not absolutely convergent. $\endgroup$
    – Simon S
    Apr 14, 2015 at 23:49
  • $\begingroup$ 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 $n>N$. $\endgroup$ Apr 14, 2015 at 23:50
  • $\begingroup$ @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! $\endgroup$ Jun 12, 2022 at 15:22

1 Answer 1

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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 $n > N$.

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