Question

Let a_n be the n th digit after the decimal point in
9pi+5e. Evaluate the sum n=1 to infinity of a_n(.1)^n.

Answer 1

I believe this should just be equal to \[9\pi+5e -\left(\left\lfloor 9\pi +5e \right\rfloor\right)=9\pi+5e-41\]The first element in your sum should be the first digit after the decimal point times \(.1\), so you have \[.8\]The next element in the sum is the second digit after the decimal place times \(.01\), so you get \[.8+.06=.86\]And so on.

Answer 2

Essentially, notice that if \(a_n\) is the \(n\)-th digit after the decimal, then \(9\pi+5e\) is equal to \[a_n(.1)^n\]Plus a bunch of other decimals and 41.

Answer 3

Thanks :-)

Answer 4

You're welcome.