Statements equivalent to $\sum\limits_{n=1}^{\infty} b_n$ diverging. Let $b_n$ be a real sequence. I'm trying to figure out if $\sum\limits_{n=1}^{\infty} b_n$ diverging is equivalent to the statement $\forall M\in\mathbb R, \forall\varepsilon>0,\ \exists n_0\in\mathbb N\text{ such that }\forall n\in\mathbb N,\ n\geq n_0\implies \left|\sum\limits_{r=1}^nb_r-M\right|\geq\varepsilon$. Certainly if the statement holds then $\sum\limits_{n=1}^{\infty} b_n$ diverges. However, I'm not entirely sure if the converse is true or not. I'm inclined to say no since if I know that $\sum\limits_{n=1}^{\infty} b_n$ diverges then by definition I have that 
$$\forall M\in\mathbb R, \exists\varepsilon>0,\ \exists n_0\in\mathbb N\text{ such that }\forall n\in\mathbb N,\ n\geq n_0\implies \left|\sum\limits_{r=1}^nb_r-M\right|\geq\varepsilon$$
ie not the given statement. However, in the same time 
my intuition tells me that if $\sum\limits_{n=1}^{\infty} b_n$ diverges then the partial sums are "blowing up", and so if you give me an arbitrary "limit point" $M \in \mathbb R$ and an arbitrary neighbourhood around it $\varepsilon>0$ then my partial sums are going to eventually move "infinitely" away from that neighbourhood.
 A: Your claim is false. Take for instance $b_n=(-1)^{n+1}$. Then, the partial sums are: 
$s_n = \sum_{j=1}^nb_j=1$ if $n$ is odd, and $s_n= 0$ if $n$ is even. This of course means that the series does not converge. Furthermore, for $M=0$, $\forall n_0 \in \mathbb{N}$, there exists an $n\geq n_0$ such that $s_n=0.$ Then, it follows that the statements are not equivalent.
If you are looking for a way to say that a series of positive numbers diverges, it is probably better to state this as: $\forall M \in \mathbb{R}$, there is an $n_0\in\mathbb{N}$ such that $\sum_{i=1}^{n}b_j \geq M$ for all $n\geq N$. This is equivalent to state that the partial sums, which form in increasing sequence or real numbers, are unbounded. This is the only possible way for such series to diverge, since bounded monotone sequences are always convergent.
A: Your intuition does not apply to series whose partial sums are bounded, but nevertheless do not converge to a limit, such as $\sum (-1)^n$.
If you restrict yourself to series with strictly positive terms, then this is true since, as you note, a divergent positive series has the property that its partial sums are arbitrarily large.
