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

MATH101 April 2018

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Question 09 (ii)
Harder: Consider the infinite series $\displaystyle \sum _{n=1}^{\infty }{\frac {\left({\sqrt {1+n}}-{\sqrt {n}}\right)}{n^{p}}}\,.$ Find, with a detailed explanation, the range of ${\textstyle p}$ for which the infinite series converges. Hint: One approach to this problem is to think about the numerator for large ${\textstyle n}$ .

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Use the fact that ${\textstyle ({\sqrt {1+n}}-{\sqrt {n}})({\sqrt {1+n}}+{\sqrt {n}})=1}$ .

Hint 2
Use the comparison test with the ${\textstyle p}$ -series.

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Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 ${\textstyle {\color {blue}{\text{the series converges if and only if }}p>1/2}}$ .