Find the limit of (-1)^n / n as n approaches infinity.

Step-by-step solution

1. 1

   Separate the bounded part from the vanishing part. The numerator $(-1)^{n}$ has no limit at all — it alternates $-1,1,-1,1,\dots$ forever. The denominator $n$ grows without bound. A bounded quantity divided by one that tends to infinity must tend to zero, and that is the whole argument in words.

2. 2

   Take absolute values to make the argument precise. Since $\left|(-1)^{n}\right|=1$ for every $n$,

   $\left|\frac{(-1)^{n}}{n}\right|=\frac{1}{n}$

   The oscillation disappears entirely once you measure size instead of sign.

3. 3

   Squeeze the sequence. For all $n\ge 1$,

   $-\frac{1}{n}\le\frac{(-1)^{n}}{n}\le\frac{1}{n}$

   and both outer bounds tend to $0$, so the trapped sequence does too.

4. 4

   State the limit.

   $\lim_{n\to\infty}\frac{(-1)^{n}}{n}=0$

   The sequence converges even though it never settles on one side of $0$: convergence requires the terms to approach a value, not to approach it from a fixed direction.

5. 5

   Note the contrast with $(-1)^{n}$ alone. Without the $\tfrac1n$ factor the sequence would have no limit, since its terms stay a distance $2$ apart forever. The vanishing factor is what rescues convergence — and it is also why the alternating harmonic series $\sum\frac{(-1)^n}{n}$ converges, by the alternating series test.