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!
 A: 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.
A: 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.
A: 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.
A: 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.
A: 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 $a_{2N}=-\dfrac{N}{3N+1} <-\dfrac{N}{3N+N} = -\dfrac{N}{4N} = -\dfrac{1}{4} \not > -\dfrac{1}{5}$.
Contradiction.
Case $2$: $L<0$. Apply similar thinking.
