Verified answer

Just look through the common taylor series

http://en.wikipedia.org/wiki/Taylor_series

find the closest one and make substitutions/modifications till it matches... In this case it's -ln(1-x)

1.3831801029685057563253819901297764612018109135177526859115...

=

1 - ln(1-1/pi)

???=990 and you can get more digits at

http://www.wolframalpha.com/input/?i=1-ln%281-1%2F...

Since 1/(1-x) = 1 + x + x^2 + x³ + ..., integrating both sides from 0 to x yields

-ln |1 - x| = x + x²/2 + x³/3 + ...

==> ln|1/(1-x)| = x + x²/2 + x³/3 + ... .

Since the geometric series converges for |x| < 1, so does the series of ln|1/(1 - x)|. Now, let x = 1/π < 1:

ln|1/(1 - (1/π))| = (1/π) + (1/π)²/2 + (1/π)³ /3 + ...

==> 1 + ln(π / (π - 1)) = 1 + 1/π + 1/(2π²) + 1/(3π³)+ ....

I hope that helps!

you need to find the common ratio. the formula for the sum of an infinite geometric series is the first term divided by 1 - the common ratio. assuming this is a geometric series.

It would not converge. making use of parentheses, we are able to make the sum seem to be distinctive values. (one million-one million)+(one million-one million)+(one million-one million)+... = 0+0+0+... = 0. one million+(-one million+one million)+(-one million+one million)+(-one million+one million)+... = one million+0+0+0... = one million. Or my very own favourite: Subtract one million from the two facets. S-one million = -one million+one million-one million+one million-... = -S. resolve for S to get S = one million/2. So yeah, there's no sum.