Science:Math Exam Resources/Courses/MATH101/April 2016/Question 05 (a)

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MATH101 April 2016

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Question 05 (a)
Does ${\displaystyle \sum _{n=1}^{\infty }{\frac {e^{-{\sqrt {n}}}}{\sqrt {n}}}}$ converge or diverge? Justify your conclusion.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Can you apply the integral test?

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Solution
Set $f(x)={\frac {e^{-{\sqrt {x}}}}{\sqrt {x}}}$ . For $x\geq 1$ , this function is positive and decreasing (since it is the product of the two decreasing functions $e^{-{\sqrt {x}}}$ and ${\frac {1}{\sqrt {x}}}$ ). Using the substitution $u={\sqrt {x}}$ , so that $du={\frac {1}{2{\sqrt {x}}}}\,dx$ , we see that ${\begin{aligned}\int _{1}^{\infty }f(x)dx&=\lim _{R\to \infty }\int _{1}^{R}\left({\frac {e^{-{\sqrt {x}}}}{\sqrt {x}}}dx\right)\\&=\lim _{R\to \infty }\left(\int _{1}^{\sqrt {R}}e^{-u}\cdot 2\,du\right)\\&=\lim _{R\to \infty }\left({-}2e^{-u}{\Big \|}_{1}^{\sqrt {R}}\right)\\&=\lim _{R\to \infty }\left({-}2e^{-{\sqrt {R}}}+2e^{-1}\right)=0+2e^{-1},\end{aligned}}$ and so this improper integral converges. By the Integral Test, the given series also converges.