Sum of $(-1)$ to the number of prime factors bounded? Define a function $f:\mathbb{N}\to\mathbb{Z}$ to be $f(1) = 1, f(p) = -1$ for all prime p, and for all $x,y\in\mathbb{N},f(xy)=f(x)f(y)$.
Try to prove that $\sum_{i=1}^{\infty}f(n)$ is not bounded.
 A: If $\sum_0^\infty a_n$ converges, then $a_n$ must tend to $0$. Since $f(n)$ clearly does not tend to $0$, your sum doesn't converge.
A: See page http://en.wikipedia.org/wiki/Liouville_function 
$f(x) = \lambda(x)$, Liouville's lambda function.
$\sum_{k=1}^n\lambda(k) = L(n)$ was considered by Pólya.  We see the statement: $L(n) > 0.06\;\sqrt{n}$ for infinitely many $n$.  This certainly shows that the series is not bounded.
A: The result cited on Wikipedia refers to a paper on oscillation theorems by Anderson and Stark.  I'm not sure there is a simpler way to prove this than using the analyticity of the Dirichlet series $\sum_{n=1}^\infty \lambda(n)/n^s = \zeta(2s)/\zeta(s)$.
If the partial sums of $\lambda(n)$ are bounded, then the Dirichlet series converges for $\Re s > 0$ and can be bounded nicely in terms of $s$ using Abel summation.  However, $\zeta(2s)/\zeta(s)$ has a pole at $s=1/2$.  An identical argument shows that $L(n) = \sum_{k=1}^n \lambda(n)$ cannot be bounded by any $O(n^{1/2 - \epsilon})$ function.
