Product of a divergent limit and a convergent limit approaching infinity I am studying convergent and divergent series and therefore have to review some limits. While doing a practice problem I encountered the follwoing problem:
$$\lim_{\ n \rightarrow \infty } \frac {{(-1)}^{n}\sqrt{n+1}}{n}$$
In this case, the ${(-1)^n}$ diverges according to Symbolab. However, the other bit converges to $0$.
$$\lim_{\ n \rightarrow \infty }\frac {\sqrt{n+1}}{n} = \lim_{\ n \rightarrow \infty }\sqrt {\frac {{n+1}}{n^2}}=\lim_{\ n \rightarrow \infty }\sqrt {\frac1n+\frac1{n^2}}=0$$
My assumption is that if we were to split the first part of the equation into two options we could say ${-1}^n$ is either negative or positive depending if $n$ is odd or even. In any case, we would still be multiplying by $0$ making the limit of the whole function $0$.
Is my line of thought wrong? And if so, why cannot I analyze the function that way?
My guess would be that I am treating $\infty$ like a number so my analysis would be invalid.

The limit is $0$, because, for each $n\in\Bbb N$,$$-\frac{\sqrt{n+1}}n\leqslant\frac{(-1)^n\sqrt{n+1}}n\leqslant\frac{\sqrt{n+1}}n$$and therefore, by the squeeze theorem, and since$$\lim_{n\to\infty}-\frac{\sqrt{n+1}}n=\lim_{n\to\infty}\frac{\sqrt{n+1}}n=0,$$the limit of your sequence is $0$.
Concerning your first approach, I don't know what you are talking about when you mention “we cannot multiply a number by $\infty$”. There is no multiplication by $\infty$ here.
