Series sum of Fractional Fibonacci Series

Question

Please help me proving that

$\sum_{n=1}^{\infty} F(n)/(10^{-(n+1)}) = 0.011235955...$

Where F(n) are the Fibonacci Numbers

Answer 1

The first terms of the summation are

$$0.01+0.001+0.0002+0.00003+0.000005+0.0000008+0.00000013+0.000000021+0.0000000034+0.00000000055+0.000000000089+0.0000000000144+0.00000000000233=\color{green}{0.011235955}05573.$$

Then it is no more possible to have a carry on the last digit $5$ and the nine first decimals are exact.

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For rigor, we can check that $F_n<2^n$. Indeed by induction

$$F_{n+2}=F_{n+1}+F_n<2^n+2^{n+1}<2^{n+2}$$ and $F_0<2^0$.

So the error in the above evaluation is less than

$$\frac4{5^{13}}=6.6\cdot10^{-10}.$$