anonymous
  • anonymous
Infinite series e^(-n)sin(n) ? what test should I use?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
Ok. Lets do this. e^(-n) can be rewritten as 1/e^n. So when we combine this with our sin(n) we get \[\sin (n)/e^n\]
anonymous
  • anonymous
I think you will need to use the Ratio Test.
anonymous
  • anonymous
Looks like a transient term, the series will converge. e^-n will be a very small number Integral test will conclude that it is a converging series.

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