Science:Math Exam Resources/Courses/MATH101/April 2018/Question 09 (ii)/Solution 1

Observe that for any integer ${\textstyle n}$ we have

$$({\sqrt {1+n}}-{\sqrt {n}})({\sqrt {1+n}}+{\sqrt {n}})=(1+n)-n=1.$$

Thus we can rewrite the summand as

$${\frac {{\sqrt {n+1}}-{\sqrt {n}}}{n^{p}}}={\frac {1}{({\sqrt {n+1}}+{\sqrt {n}})n^{p}}}.$$

We will now apply the comparison test with the ${\textstyle p}$-series.

For every positive integer ${\textstyle n}$ we have

$${\Bigg |}{\frac {1}{({\sqrt {n+1}}+{\sqrt {n}})n^{p}}}{\Bigg |}\leq {\frac {1}{{\sqrt {n}}\cdot n^{p}}}={\frac {1}{n^{p+{\frac {1}{2}}}}}$$

so the series converges if ${\textstyle p>1/2}$. On the other hand, every positive integer ${\textstyle n}$ satisfies

$${\sqrt {1+n}}+{\sqrt {n}}\leq 3{\sqrt {n}},$$

so we also have

$${\frac {1}{({\sqrt {1+n}}+{\sqrt {n}})n^{p}}}\geq {\frac {1}{3{\sqrt {n}}\cdot n^{p}}}={\frac {1}{3n^{p+{\frac {1}{2}}}}}.$$

This shows that the series diverges to ${\textstyle +\infty }$ when ${\textstyle p\leq 1/2}$.

Answer: The correct answer is that ${\textstyle {\color {blue}{\text{the series converges if and only if }}p>1/2}}$.