Zeros & Nines

Calculus Level 3

( 10 9 × 99 + 100 99 × 999 + 1000 999 × 9999 + ⋯ ) − 1 ? \left(\dfrac{10}{9\times99} + \dfrac{100}{99\times999} + \dfrac{1000}{999\times9999} + \cdots \right)^{-1} \, ?

1 9 81 27 3

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Ar Hakim by AR Hakim , 25, Indonesia

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1 solution

Shaun Leong
Jan 24, 2016

∑ n = 1 ∞ 1 0 n ( 1 0 n − 1 ) ( 1 0 n + 1 − 1 ) \displaystyle \sum_{n=1}^\infty \dfrac {10^n}{(10^n-1)(10^{n+1}-1)} = ∑ n = 1 ∞ 1 9 ( 1 1 0 n − 1 − 1 1 0 n + 1 − 1 ) =\displaystyle \sum_{n=1}^\infty \dfrac {1}{9}(\dfrac {1}{10^n-1} - \dfrac {1}{10^{n+1}-1}) = 1 9 ∑ n = 1 ∞ 1 1 0 n − 1 − ∑ n = 2 ∞ 1 1 0 n − 1 =\dfrac {1}{9} \displaystyle \sum_{n=1}^\infty \dfrac {1}{10^n-1} - \displaystyle \sum_{n=2}^\infty \dfrac {1}{10^n-1} = 1 9 ∗ 1 10 − 1 =\dfrac {1}{9}*\dfrac {1}{10-1} = 1 81 =\dfrac {1}{81} ( 1 81 ) − 1 = 81 (\dfrac {1}{81})^{-1}=\boxed{81}

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