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Q.
The limiting sum of the infinite series, $\frac{1}{10}+\frac{2}{10^2}+\frac{3}{10^3}+\ldots$. whose $n ^{\text {th }}$ term is $\frac{ n }{10^{ n }}$ is equal to
Sequences and Series
Solution:
Let $S =\frac{1}{10}+\frac{2}{10^2}+\frac{3}{10^3}+\ldots \ldots \ldots . .$(1)
So, $ \frac{ S }{10}=0+\frac{1}{10^2}+\frac{2}{10^3}+\ldots \ldots \ldots$(2)
$\therefore $ Subtracting (2) from (1), we get
$ \frac{9 S }{10}=\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\ldots \ldots \ldots . . .=\frac{1 / 10}{1-\frac{1}{10}}=\frac{1}{9} $
$\Rightarrow S = \frac{1}{9}\left(\frac{10}{9}\right)=\frac{10}{81} $