 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/19/a/iv/solution

For the positive terms

u_{n} = 2^{n} / n!

, the consecutive ratio is

u_{n + 1} / u_{n} = 2/ (n + 1) \to 0

, so the ratio test proves convergence. The exponential series gives

e^{2} = \sum_{n = 0}^{\infty} 2^{n} / n!

. Removing the

n = 0

term yields

n = 1 \sum \infty \frac{2 ^{n}}{n !} = e^{2} - 1.

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