Science:Math Exam Resources/Courses/MATH101 C/April 2024/Question 20 (c)

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MATH101 C April 2024

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Other MATH101 C Exams

• April 2024 •

Question 20 (c)
Let $f(x)=\int _{0}^{x}te^{-t}\;\mathrm {d} t.$ Use this function in all parts below. (c) For the constant c found in part (b), find the exact value of S.

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
Recognize that the given series resembles a power series of the form $\sum (n+1)x^{n}$ . Try starting with the basic geometric series to obtain a formula for the given sum.

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Solution
In part b) we found the constant to be $c=1$ . Thus, we seek the exact value of $S=\sum _{n=0}^{\infty }(n+1)e^{n}.$ This is related to the function ${\begin{aligned}f(x)=\sum _{n=0}^{\infty }(n+1)x^{n},\end{aligned}}$ at the value at $x=e^{-1}.$ To evaluate it, we use known power series results. We recall the sum of a geometric series: ${\begin{aligned}\sum _{n=0}^{\infty }x^{n}={\frac {1}{1-x}},\quad {\rm {for}}\;\|x\|<1.\end{aligned}}$ To generate the term $(n+1)x^{n}$ , differentiate both sides of the geometric series with respect to $x$ ${\begin{aligned}f(x)=\sum _{n=0}^{\infty }(n+1)x^{n}={\frac {d}{dx}}{\frac {1}{1-x}}={\frac {1}{(1-x)^{2}}},\quad x\in (-1,1).\end{aligned}}$ The point $x=e^{-1}$ lies in the interval of convergence, so we can substitute it ${\begin{aligned}S=f(e^{-1})={\frac {1}{(1-e^{-1})^{2}}}={\frac {e^{2}}{(e-1)^{2}}},\end{aligned}}$ obtaining the exact value of $S.$