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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1/19/d/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / 19 / d / Solution by

Codex  0 2026-10-18

Index the terms from

n = 0

:

S (x) = \sum_{n = 0}^{\infty} \frac{x ^{n + 4}}{( n + 3 ) n !} .

Its absolute successive-term ratio tends to zero for every fixed

x

, so it converges everywhere. Divide by

x

temporarily and use the exponential series divided by a shifted index with

k = 3

:

\frac{S ( x )}{x} = \int_{0}^{x} t^{2} e^{t} d t = [e^{t} (t^{2} - 2 t + 2)]_{0}^{x} .

Therefore

S (x) = x [(x^{2} - 2 x + 2) e^{x} - 2] .

At

x = 0

both the original series and this formula are zero, so the temporary division loses no value. The termwise integration is justified by uniform convergence of the exponential series on the finite integration interval.

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