 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/19/a/solution

For

r \neq = 1

, differentiate the finite geometric series and multiply by

r

\sum_{n = 1}^{N} n r^{n} = r \frac{d}{d r} (\frac{1 - r ^{N + 1}}{1 - r}) = \frac{r [ 1 - ( N + 1 ) r ^{N} + N r ^{N + 1} ]}{( 1 - r ) ^{2}} .

The required terms are

- n r^{n}

with

r = - 1/3

S_{N} = \frac{3}{16} [1 - (1 + \frac{4 N}{3}) (- \frac{1}{3})^{N}] .

Since

N / 3^{N} \to 0

, the correction tends to zero, proving convergence and yielding

n = 1 \sum \infty \frac{( - 1 ) ^{n + 1} n}{3 ^{n}} = \frac{3}{16} .

In fact the ratio test also proves absolute convergence, since the ratio of consecutive absolute terms tends to

1/3

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