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jmc

algebra senior

Problem

Let

f (r) = \sum_{j = 2}^{2008} \frac{1}{j ^{r}} = \frac{1}{2 ^{r}} + \frac{1}{3 ^{r}} + \dots + \frac{1}{200 8 ^{r}}

. Find

\sum_{k = 2}^{\infty} f (k)

.

Solution

We change the order of summation:

k = 2 \sum \infty j = 2 \sum 2008 \frac{1}{j ^{k}} = j = 2 \sum 2008 k = 2 \sum \infty \frac{1}{j ^{k}} = j = 2 \sum 2008 \frac{1}{j ^{2} ( 1 - \frac{1}{j} )} = j = 2 \sum 2008 \frac{1}{j ( j - 1 )} = j = 2 \sum 2008 (\frac{1}{j - 1} - \frac{1}{j}) = 1 - \frac{1}{2008} = \frac{2007}{2008} .

Final answer

\frac{2007}{2008}