Summation of Exponential function with mod from negative infinity to infinity

Question

Can someone explain how this is true:-

$$ \sum_{k = -\infty}^{\infty} e^{-|k|} = -1 + \sum_{k = -\infty}^{0} e^{-|k|} + \sum_{k=0}^{\infty} e^{-|k|} $$

I cannot understand why the negative 1 comes in the end of the expression.

Answer 1

In general $$\sum_{k = - \infty}^{\infty} f(k) = \sum_{k = - \infty}^{-1} f(k) + f(0) + \sum_{k=1}^{\infty} f(k)$$ and by setting $k$ to $-k$ in the first summation it reduces to $$\sum_{k = - \infty}^{\infty} f(k) = f(0) + \sum_{k=1}^{\infty} (f(k) + f(-k)).$$ If the summation is to begin with $k=0$ then: \begin{align} \sum_{k = - \infty}^{\infty} f(k) &= f(0) + \sum_{k=1}^{\infty} (f(k) + f(-k)) + (f(0) + f(-0)) - (f(0) + f(-0)) \\ &= - f(-0) + \sum_{k=0}^{\infty} (f(k) + f(-k)) \\ &= - f(0) + \sum_{k=1}^{\infty} (f(k) + f(-k)). \end{align}

In the case of $f(k) = e^{-|k|}$ then $$\sum_{k = - \infty}^{\infty} e^{-|k|} = e^{-|0|} + 2 \, \sum_{k=1}^{\infty} e^{-|k|} = 1 + 2 \, \sum_{k=1}^{\infty} e^{-|k|}.$$

Now, if the summation is to start at $k=0$ then $$\sum_{k = - \infty}^{\infty} e^{-|k|} = -1 + 2 \, \sum_{k=0}^{\infty} e^{-|k|}.$$