Can I do this rearrangement with an infinite series to arrive at divergent parts? $$\sum_{i=3}^{\infty}\frac{i}{\left(i-2\right)\left(i-1\right)\left(i+1\right)} = \sum_{i=3}^\infty \frac{2}{3\left(i-2\right)} - \frac{1}{2\left(i-1\right)} - \frac{1}{6\left(i+1\right)} \\**= \frac{2}{3}\sum_{i=1}^\infty\frac{1}{i} - \frac{1}{2}\left(\left(\sum_{i=1}^\infty\frac{1}{i}\right) - 1\right) - \frac{1}{6}\left(\left(\sum_{i=1}^\infty\frac{1}{i}\right) - 1 - \frac{1}{2} - \frac{1}{3}\right)** \\= \left(\frac{2}{3} - \frac{1}{2} - \frac{1}{6}\right) \left(\sum_{i=1}^\infty\frac{1}{i}\right) + \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{18} \\= 0\cdot\left(\sum_{i=1}^\infty\frac{1}{i}\right) + \frac{29}{36} \\= \frac{29}{36} \space or \space \infty?$$
When summing up the series above I came by the harmonic series multiplied by $0$. I simply equated it to be $0$ while my friend keeps arguing that the harmonic series is an infinity and infinity multiplied by $0$ is undefined. Now I find it difficult to argue with it. Does this mean that the series above diverge?
Or maybe I was thinking, in the starred line, the $\frac{2}{3}\left(\sum_{i=1}^\infty\frac{1}{i}\right) - \frac{1}{2}\left(\sum_{i=1}^\infty\frac{1}{i}\right) - \frac{1}{6}\left(\sum_{i=1}^\infty\frac{1}{i}\right)$ is cancelling out anyway, right? I mean I could just right it term by term and show every term cancels. If they are cancelling out then this rearrangement should not be problematic, right? So, am I right to say that $0\cdot\left(\sum_{i=1}^\infty\frac{1}{i}\right) = 0$ to show it's convergence?
 A: If the series begins from $i=3$, it will converge, but you need to be more careful with the manipulation of the series, you can't just split the series in three. The theorem that would allow you to split series doesn't apply here:
$$\sum_n (a_n - b_n) = \sum_n a_n - \sum_n b_n$$
IF $ \sum_n a_n$ AND $\sum_n b_n$ CONVERGE.
A: Well you can use this method for verification
$\displaystyle\sum_{i=3}^{\infty}\frac{i}{\left(i-2\right)\left(i-1\right)\left(i+1\right)}$
$\displaystyle\sum_{i=3}^{\infty}\frac{i^2}{\left(i-2\right)\left(i-1\right)(i)\left(i+1\right)}$
$\displaystyle\sum_{i=3}^{\infty}\frac{(i-1)(i+1)+1}{\left(i-2\right)\left(i-1\right)(i)\left(i+1\right)}$
$\displaystyle\sum_{i=3}^{\infty}\frac{(i+1)(i-1)}{\left(i-2\right)\left(i-1\right)(i)\left(i+1\right)}+\sum_{i=3}^{\infty}\frac{1}{\left(i-2\right)\left(i-1\right)(i)\left(i+1\right)}$
$\displaystyle\sum_{i=3}^{\infty}\frac{1}{\left(i-2\right)(i)}+\sum_{i=3}^{\infty}\frac{1}{\left(i-2\right)\left(i-1\right)(i)\left(i+1\right)}$
$\displaystyle\frac{1}{2}\sum_{i=3}^{\infty}\frac{i-(i-2)}{\left(i-2\right)(i)}+\frac{1}{3}\sum_{i=3}^{\infty}\frac{(i+1)-(i-2)}{\left(i-2\right)\left(i-1\right)(i)\left(i+1\right)}$
$\displaystyle\frac{1}{2}\sum_{i=3}^{\infty}(\frac{1}{\left(i-2\right)}-\frac{1}{i})+\frac{1}{3}\sum_{i=3}^{\infty}(\frac{1}{\left(i-2\right)\left(i-1\right)(i)}-\frac{1}{\left(i-1\right)\left(i\right)(i+1)})$
$\displaystyle\frac{1}{2}(1+\frac{1}{2})+\frac{1}{3}(\frac{1}{6})$
$\displaystyle\frac{3}{4}+\frac{1}{18}={\frac{29}{36}}$
A: UPDATE: This question was radically rewritten. I was answering the original question: What is 0 multiplied with a divergent series with summation of infinity?
The series (the fixed variant) converges, its sum is $\frac{29}{36}.$
As to the separate question of zero being multiplied by a divergent series, the main algebraic property of zero is that it multiplied by anything is still zero, and this is usually taken as the stronger rule than multiplication by something infinite. Even in the extended real line ${\overline {\mathbb {R} }}$ while $0\cdot\infty$ is often left undefined, in many cases there is a solid justification to define it as $0$.
