$$\sum\frac{(-1)^n}{\sqrt{n+1}} \text{and} \sum\frac{1}{\sqrt{n+1}}$$
The first one is an alternating series, so it would just be:
$$\sum (-1)^n\frac{1}{\sqrt{n+1}}\Rightarrow \;^\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n+1}} \Rightarrow \frac{\frac1n}{\sqrt{1+\frac1n}}\Rightarrow \frac{0}{1}=0 = \text{convergent}$$
But for the second one I'm confused a little:
$$^\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n+1}} \Rightarrow \frac{\frac1n}{\sqrt{1+\frac1n}}\Rightarrow \frac{0}{1}=0 = \text{convergent}$$
or
$$\sum\frac{1}{\sqrt{n+1}} \lt \frac{1}{\sqrt{n}}\Rightarrow\frac{1}{n^{1/2}} = \text{p-series}\frac12\lt1 = \text{divergent}$$
Is the second series convergent or divergent, and by what test(s)?
Note: in my last example I am comparing the series to a greater/divergent series
Best Answer
You are making a big mistake. The fact that $$\lim\limits_{n\rightarrow \infty}a_n = 0 $$ does not imply that $\sum a_n$ is convergent. On the other hand, if the limit is nonzero, we can conclude that the sum is divergent.
The first series is an alternating series and converges by the Leibniz alternating series test. For the second series compare $$\frac{1}{\sqrt{n+1}}\geq \frac 1n $$ when $n\geq 2$ and $\sum\frac1n$ is the classic example of a divergent series, therefore so is the second series.
ADDED: Since you're confused about the p-test. $$\sum\limits_{n=1}^\infty \frac{1}{\sqrt{n+1}} = \sum\limits_{n=2}^\infty \frac{1}{\sqrt{n}} = \sum\limits_{n=2}^\infty \frac{1}{n^{1/2}}$$ hence the series diverges by the p-test.