Science:Math Exam Resources/Courses/MATH101 B/April 2025/Question 08 (b)

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MATH101 B April 2025

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  •  April 2024  •  April 2025  •  

Question 08 (b)
In this question, $a$ is a real number. Find, with justification, all values of $a$ for which the series $$\begin{array}{r}\sum _{n=1}^{\mathrm{\infty }}n{e}^{-an}.\end{array}$$ converges.

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

First consider the general term $a_n=n{e}^{-an}$. What happens to $\underset{n\to \mathrm{\infty }}{\lim }n{e}^{-an}$ when $a\le 0$? For $a>0$, try the ratio test: $$\frac{a_{n+1}}{a_n}=\frac{(n+1)e^{-a(n+1)}}{n{e}^{-an}}.$$ For which values of $a$ is this limit less than 1? $(\text{Alternatively, you may compare with the improper integral}\underset{1}{\overset{\mathrm{\infty }}{\int }}x{e}^{-ax}dx.)$

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Solution

For this question, we consider $a\le 0$ and $a>0$ seperately. If $a\le 0$, $\underset{n\to \mathrm{\infty }}{\lim }n{e}^{-an}=+\mathrm{\infty }$, so the series diverges by the divergence test. For $a>0$, we can use the integral test. For $x\ge 1$, $f(x)=x{e}^{-ax}$ is positive and continuous. Also, $f^{\prime }(x)=e^{-ax}-ax{e}^{-ax}=(1-ax)e^{-ax}<0\text{for }x>\frac{1}{a},$ so $f$ is eventually decreasing. By integration by parts, the improper integral is $$\underset{1}{\overset{\mathrm{\infty }}{\int }}x{e}^{-ax}dx=\underset{b\to \mathrm{\infty }}{\lim }\underset{1}{\overset{b}{\int }}x{e}^{-ax}dx=\underset{b\to \mathrm{\infty }}{\lim }[-\frac{1}{a}x{e}^{-ax}{|}_1^b+\frac{1}{a}\underset{1}{\overset{b}{\int }}{e}^{-ax}dx].$$ Compute: $$=\underset{b\to \mathrm{\infty }}{\lim }[-\frac{b}{a}{e}^{-ab}+\frac{1}{a}{e}^{-a}-\frac{1}{a^2}{e}^{-ax}{|}_1^b]$$ $$=\underset{b\to \mathrm{\infty }}{\lim }[-\frac{b}{a}{e}^{-ab}+\frac{1}{a}{e}^{-a}-\frac{1}{a^2}{e}^{-ab}+\frac{1}{a^2}{e}^{-a}]$$ $$=(\frac{1}{a}+\frac{1}{a^2})e^{-a}.$$ Thus the integral converges. By the integral test, the series also converges. Hence, the series converges if and only if $a>0$. Alternate solution uses the ratio test: $$\underset{n\to \mathrm{\infty }}{\lim }\left|\frac{(n+1)e^{-a(n+1)}}{n{e}^{-an}}\right|=e^{-a}.$$ Thus the series converges for $a>0$ and diverges for $a<0$. For $a=0$, the series becomes $\sum n$, which diverges. Therefore, the series converges if and only if $a>0$.

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