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.

I believe this should just be equal to
9π+5e−(⌊9π+5e⌋)=9π+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

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