Calculating $\sum_{k=1}^\infty 2^{-k}(e^{-k}-e^{-k-1})$

Question

Pretty basic question - I'm just not that experienced with calculating summations and would love help with understanding the steps involved (I computed it with Mathematica but couldn't see the steps there). I'm not sure if I can distribute the sum, or if I can coerce this to use the Taylor series for $e^x$. Thanks!

Answer 1

Hint:

$$ \sum_{k=1}^\infty 2^{-k}(e^{-k}-e^{-k-1}) = \sum_{k=1}^{\infty} (2e)^{-k} \left( 1- e^{-1}\right)= \left(1 - \frac{1}{e}\right) \sum_{k=1}^{\infty} \frac{1}{(2e)^k}. $$

Can you finish the problem from here?