# Why $(-1)^{n-1} \frac{n}{3n + 1}$ is divergent?

I am wondering the reason $$(-1)^{n-1} \frac{n}{3n + 1}$$ is divergent although the limit of $$\frac{n}{3n + 1}$$ is $$1/3.$$

Any explanation will be greatly appreciated!

• @DonThousand: The sequence can't converge since the limit is not $0$ and it does not have an ultimately constant sign. Jun 17, 2021 at 20:48
• A sequence is divergent if it doesn't converge. Since $n/(3n+1)$ converges to $1/3$, your sequence looks like $1/3,-1/3,1/3,-1/3,...$, which is osscilating.
– pax
Jun 17, 2021 at 20:49

Let $$a_n = (-1)^{n-1} \frac{n}{3n+1}$$. Then $$a_{2n} = -\frac{2n}{6n+1}$$ and $$a_{2n+1} = \frac{2n+1}{6n+4}$$. Observe that $$\lim_{n\to \infty}a_{2n} = -\frac{1}{3} \qquad \lim_{n\to \infty}a_{2n+1} = \frac{1}{3}$$

Since the limits along two subsequences differ, then $$a_n$$ diverges.

• Hi, what is the criterium that have you used? At this moment I have forgotten it. +1 Jun 17, 2021 at 21:00
• @Sebastiano If $x_n \to a$ , then any subsequence satisfies $x_{n_k} \to a$. If the conclusion does not hold for a sequence, by contrapositive it must be that the parent sequence does not converge. Jun 17, 2021 at 21:01
• Thank you very much for your collaboration....thank you again. Jun 17, 2021 at 21:04

Consider two subsequences of the given sequence, viz. the even subsequence, and the odd subsequence.

• For even $$n$$, we may write $$n=2k$$ for $$k\in\mathbb{N}$$. Then we have : $$\lim_{n\to\infty}(-1)^{n-1} \frac{n}{3n + 1} = \lim_{k\to\infty}(-1)^{2k-1} \frac{2k}{6k + 1} = -\lim_{k\to\infty} \frac{1}{3 + \frac{1}{2k}} = -\frac{1}{3}$$

• For odd $$n$$, we may write $$n=2k+1$$ for $$k\in\mathbb{N}$$. Then we have : $$\lim_{n\to\infty}(-1)^{n-1} \frac{n}{3n + 1} = \lim_{k\to\infty}(-1)^{2k+1-1} \frac{2k+1}{6k + 3 + 1} = \lim_{k\to\infty} \frac{1}{3 + \frac{1}{2k+1}} = \frac{1}{3}$$

So, the odd and the even subsequences of the original sequence converge to two different limit points. Thus, the sequence is not convergent.

Convergence implies that there exists a singular limit to the sequence, which itself requires that for all $$\epsilon > 0$$, there exists an $$N \in \mathbb{N}$$ such that $$|a_n - a| < \epsilon, \quad \forall n \geq N.$$ This convergence condition does not hold true for the sequence $$a_n = \frac{(-1)^{n-1}n}{3n+1}$$ purely due to its oscillatory nature and the fact that its limit when the $$(-1)^n$$ is removed, is nonzero.

Since the sequence $$\frac{n}{3n+1}$$ converges to a nonzero value, convergence of $$(-1)^n\frac{n}{3n+1}$$ would imply convergence of $$(-1)^n.$$ However, this sequence is known to be divergent.

Suppose a limit $$L$$ exists.

Case $$1$$: $$L\ge 0$$. Then for all $$\epsilon > 0$$, there is some $$N\in\mathbb N$$ such that for all $$n\ge N$$

$$|a_n-L|<\epsilon.$$

Choose $$\epsilon = 1/5$$, and find such an $$N$$ as above. Then as $$n=2N > N$$ we must have $$|a_{2N}-L|<1/5$$ or

$$-1/5 < L-1/5 < -\dfrac{N}{3N+1}=a_{2N} < L+1/5$$

but $$a_{2N}=-\dfrac{N}{3N+1} <-\dfrac{N}{3N+N} = -\dfrac{N}{4N} = -\dfrac{1}{4} \not > -\dfrac{1}{5}$$.

Case $$2$$: $$L<0$$. Apply similar thinking.