Does $\sum_{n=1}^{\infty}\left(-1\right)^{n+1}\frac{3\sqrt{n+1}}{\sqrt{n}+1}$ Diverge or Converge?

Question

Does the following summation converge or diverge? \begin{align} \sum_{n=1}^{\infty}\left(-1\right)^{n+1}\frac{3\sqrt{n+1}}{\sqrt{n}+1}.\tag{1} \end{align} I don't know where to begin. I think I should, however, use the absolute value test s. th. if \begin{align} \sum_{n=1}^{\infty}\left|\left(-1\right)^{n+1}\frac{3\sqrt{n+1}}{\sqrt{n}+1}\right|=\sum_{n=1}^\infty \frac{3\sqrt{n+1}}{\sqrt{n}+1}=3\sum_{n=1}^{\infty}\frac{\sqrt{n+1}}{\sqrt{n}+1}\tag{2} \end{align} converges then $\left(1\right)$ converges. If I pick arbitrarily large values of $n$ then it seems $\left(2\right)$ does not get smaller and eventually $\rightarrow 0$.

Thank you for your time,

Answer 1

Use term test. Since $$ \lim_{n\to\infty}\frac{3\sqrt{n+1}}{\sqrt{n}+1}=\lim_{n\to\infty}\frac{3\sqrt{n+1}}{\sqrt{n}+1}=\lim_{n\to\infty}\frac{3\sqrt{1+\frac{1}{n}}}{1+\frac{1}{\sqrt{n}}}=3,$$ then $$ \lim_{n\to\infty}(-1)^{n+1}\frac{3\sqrt{n+1}}{\sqrt{n}+1}=3$$ does not exist because it jumps between $3$ and $-3$. So the series does not converge.

Answer 2

$$\lim_{n\to\infty}{\sqrt{n+1}\over\sqrt n +1}=\lim_{n\to\infty}{\sqrt{1+1/n}\over1+1/\sqrt n}=1.$$

Hence $$\lim_{n\to\infty}(-1)^{n+1}{3\sqrt{n+1}\over\sqrt n + 1}$$ does not exist, and the series diverges.