Finding limit of sequence quite hard Im trying to find this limit as n goes to infinity. I raised it to 1/n power bit I cant get it to a  form so I can substitute n with infinity . Form answer book I know its 0, bit I just cant see how to deal with this.
$$\lim_{ n\rightarrow\infty} (-1)^n\frac{\sqrt{1+n}}{n}$$
 A: The answer of @lmaosome does the work.  Also notice that any time you see an alternating sequence as $(-1)^n$ then the only way for $(-1)^nb_n$ to converge, it is that $b_n\to 0$. So in your case you should check that $b_n=\frac{\sqrt{n+1}}{n}\to 0$. Another way to see why this happens, $$\frac{\sqrt{n+1}}{n}=\frac{\sqrt{n+1}}{n}\cdot \frac{\sqrt{n+1}}{\sqrt{n+1}} =\frac{1}{\sqrt{n+1}}+\frac{1}{n\sqrt{n+1}}$$ since the denominators $n\sqrt{n+1}\to\infty$ and $\sqrt{n+1}\to\infty$ you get $b_n\to0$. So using this you get $(-1)^nb_n\to 0$ which in case of convergence is the only limit possible.
A: Note that $\lim_{n\rightarrow \infty} a_nb_n = \lim_{n\rightarrow\infty}a_n\lim_{n\rightarrow\infty}b_n$ if $a_n, b_n$ are convergent.
Consider $$(-1)^n\frac{\sqrt{1+n}}{n} = \frac{(-1)^n}{\sqrt{n}}\frac{\sqrt{1+n}}{\sqrt{n}}.$$ Put $a_n = \frac{(-1)^n}{\sqrt{n}}$ and $b_n = \frac{\sqrt{1+n}}{\sqrt{n}}.$ It is easy to show that both $a_n$ and $b_n$ are convergent with $\lim_{n\rightarrow\infty}a_n = 0$, and $\lim_{n\rightarrow\infty}b_n = 1$. Using the aforementioned result, $$\lim_{n\rightarrow\infty}(-1)^n\frac{\sqrt{1+n}}{n} = 0.$$
