Convergent or divergent series.

Question

Is this series convergent or divergent? Which test should one use when we are dealing with $n$th root? $$\sum_{n=0}^{\infty} \frac 1{\sqrt n+\sqrt {n+1}+\sqrt {n+2}+\sqrt {n+3}}$$

Answer 1

Consider that $$\frac{1}{\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}}\geq \frac{1}{4\sqrt{n+3}}$$ so $$\sum_{n=0}^{\infty} \frac{1}{\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}}\geq \sum_{n=0}^{\infty} \frac{1}{4\sqrt{n+3}} = \infty$$

Answer 2

For $n $ great enough, $$n <n+1 <n+2 <n+3 <n^2$$ thus $$u_n>\frac {1}{4n} $$

your series is then Divergent, by comparison test.

Answer 3

Since you ask what test should be used, another useful test is the p-series test used together with the comparison test which is an alternate explanation of what Michael Lee did in his answer.

$$ \sum_{n=1}^\infty\frac{1}{n^p}$$

converges if and only if $p>1$.

Thus

$$ \sum_{n=1}^\infty\frac{1}{\sqrt{n}} $$

diverges since $p=\dfrac{1}{2}$. The p-series test can be proved using the integral test.

The p-series test applies in this instance since

$$ \sum_{n=0}^{\infty} \frac{1}{4\sqrt{n+3}}=\frac{1}{4}\sum_{n=3}^{\infty} \frac{1}{\sqrt{n}} $$