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March 1, 2015

March 1, 2015

Posted by **sarah** on Tuesday, February 26, 2008 at 11:08pm.

the sum from k=1 to infinity of

sin(e^-k)

I'm not sure where to start..

- calculus -
**drwls**, Wednesday, February 27, 2008 at 3:06amAs k becomes large, e^-k becomes much less than 1, and sin(e^-k) approaches e^-k

The sum of the series 1 + 1/e + 1/e^2 converges to

1 /(1 - 1/e)= 1.582

High-order terms of the series

sin(e^-k) will behave similarly, but the sum of the entire series will be somethat less than 1.582.

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