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.
Best Answer
Your idea is fine since $a_n=\frac {\sqrt{n+1}}{n} \to 0$ and $b_n=(-1)^n$ is bounded by $|b_n|\le M=1$ it is a general result that by squeeze therorem
$$\left|a_n \cdot b_n \right|\le |a_n|M \to0 \cdot M =0 \implies a_n \cdot b_n \to 0$$