Let a_n be the n th digit after the decimal point in 4pi+4e.
∞
Evaluate ∑ a_n(.1)^n.
n=1
thanks for any help.
Since $\displaystyle a_n$ represents the nth digit of the decimal part of $\displaystyle 4\pi + 4e$ wouldn't it be reasonable to say that
$\displaystyle \sum_{n = 1}^{\infty} a_n(0.1)^n = (4\pi - 4e) - \lfloor 4\pi + 4e \rfloor $
in other words the decimal part of $\displaystyle 4\pi + 4e$.
As the two numbers are transcendental I would be very surprised to find that there is an exact answer other than this.
-Dan
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