<s>[INST] Why is the convergence point of $ \sum _{n=1}^{\infty }\frac{1}{2^n}-\frac{1}{2^{n+1}} $ negative? I am trying to evaluate $\frac1{2^1} - \frac1{2^2} + \frac1{2^3} - \frac1{2^4} + \cdots$
I re-wrote the sum using sigma notation as:
$$ \sum _{n=1}^{\infty } \left( \frac{1}{2^n}-\frac{1}{2^{n+1}} \right) \quad (1) $$
Hence,
$$ (1) =  \sum _{n=1}^{\infty } \frac{2^{n+1} - 2^n }{2^n2^{n+1}} = \sum _{n=1}^{\infty } \frac{2^n(2-1) }{2^n2^{n+1}} = - \sum _{n=1}^{\infty }  \frac1{2^{n+1}}  = - \sum _{n=2}^{\infty }  \frac1{2^{n}} = -\left( \sum _{n=0}^{\infty }  \left(\frac12\right)^n -1-\frac12\right) = -2+\frac32 =-\frac12 $$
Therefore,
$$ \boxed{\sum _{n=1}^{\infty }\frac{1}{2^n}-\frac{1}{2^{n+1}} =-\frac12} $$

I can't spot any algebraic mistakes, thus I assume the sum is correct. But I don't understand the result. How can a real valued sum converge to a negative number?
 [/INST] The series you are trying to evaluate is
$$\sum_{n=1}^\infty(-1)^{n-1}\left(\frac12\right)^n=\frac12\frac1{1+\dfrac12}=\frac13.$$
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