# Convergent or divergent series. [closed]

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

## closed as off-topic by Henrik, Sahiba Arora, kingW3, Shaun, Davide GiraudoAug 4 '17 at 17:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, Sahiba Arora, kingW3, Shaun, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.

• Should this be $\frac 1{\sqrt n+\sqrt {n+1}+\sqrt {n+2}+\sqrt {n+3}}$ ? But then why speak about $n^{th}$ roots? – lulu Aug 4 '17 at 15:04
• Yes that's what it is I'm so sorry this my 1st time asking question and I was unable to use the syntax of this site. – Harshal Sharma Aug 4 '17 at 15:06
• I edited your post, though I still don't understand why you are speaking of $n^{th}$ roots when there are only square roots in the expression. – lulu Aug 4 '17 at 15:10
• Quite... remember that "$n^{th}$ roots" refer to things like $\sqrt[n]{5}$ or $5^{\frac{1}{n}}$. If there are in fact n'th roots then explain what is still incorrect about the expression as it currently appears. As for syntax and writing with MathJax and $\LaTeX$, visit this page for a primer. – JMoravitz Aug 4 '17 at 15:13

## 3 Answers

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

• Sir, can you please explain the last step as how 1/4\sqrt{n+3} became ∞. I was asking which test should we use when we are dealing with square roots, cube roots etc in the question. – Harshal Sharma Aug 4 '17 at 15:12
• Well, we know that $\sum_{n=1}^{\infty} \frac{1}{n}$, the harmonic series, diverges, right? Now, consider that for $n$ big enough, $4\sqrt{n+3}$ is smaller than $n$, so $\frac{1}{4\sqrt{n+3}}\geq \frac{1}{n}$. Therefore, this series will diverge. – Michael Lee Aug 4 '17 at 15:14
• You're right, fixed. For reference, @eras'q, this is called the comparison test, although if you know the definition of convergence for an infinite series and the basic properties of limits, it should hopefully be pretty easy to see without knowing any of the other convergence tests. – Michael Lee Aug 4 '17 at 15:19

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.

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

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