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algebra intermediate

Problem

Find the sum

0.1 + 0.02 + 0.003 + 0.0004 + \dots + \frac{n}{1 0 ^{n}} + \dots .

Solution

Let

S = \frac{1}{10} + \frac{2}{1 0 ^{2}} + \frac{3}{1 0 ^{3}} + \dots .

Then

\frac{1}{10} S = \frac{1}{1 0 ^{2}} + \frac{2}{1 0 ^{3}} + \frac{3}{1 0 ^{4}} + \dots .

Subtracting these equations, get

\frac{9}{10} S = \frac{1}{10} + \frac{1}{1 0 ^{2}} + \frac{1}{1 0 ^{3}} + \dots = \frac{1/10}{1 - 1/10} = \frac{1}{9} .

Therefore,

S = \frac{10}{81} .

Final answer

\frac{10}{81}