Science:Math Exam Resources/Courses/MATH101 A/April 2024/Question 20 (b)

MATH101 A April 2024

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Question 20 (b)
Let $f(x)=\int _{0}^{x}te^{-t}\;\mathrm {d} t.$ Use this function in all parts below. (b) Given that the following series converges, find the constant c: $S=\sum _{n=0}^{\infty }(c-f(n)).$

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 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
Consider the behavior of $f(n)$ as $n\to \infty$. What must be true about $c-f(n)$ for the series $S$ to converge? Think about how the terms behave and recall the basic divergence test, which states that a series $\sum a_{n}$ can only converge if $a_{n}\to 0$.

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Solution

The Basic Divergence Detector rejects the series instantly, unless $${\begin{aligned}0=\lim _{n\to \infty }[c-f(n)]=\lim _{n\to \infty }[c-1+(n+1)e^{-n}].\end{aligned}}$$ We can compute the following limit $${\begin{aligned}\lim _{n\to \infty }(n+1)e^{-n}=\lim _{n\to \infty }{\frac {n+1}{e^{n}}},\end{aligned}}$$ by applying L'Hôpital’s Rule $${\begin{aligned}\lim _{n\to \infty }{\frac {1}{e^{n}}}=0.\end{aligned}}$$ Then the only way for the following expression to hold true, $${\begin{aligned}0=\lim _{n\to \infty }[c-1]\end{aligned}}$$ is if $c=1.$