Why is infinity multiplied by zero considered zero here?

Solution 1:

$\sum_{n=1}^{\infty}a_{n}$ is formally the limit $\lim_{n\rightarrow\infty}s_{n}$ where $s_{n}=\sum_{k=1}^{n}a_{k}$.

In the case you mention ($x=0$) we have $s_{n}=0$ for each $n$, hence $\lim_{n\rightarrow\infty}s_{n}=0$

Solution 2:

It is worth noting that although $+\infty\times0$ is an indeterminate form (which is a statement about the limit of a product expressions where one factor tends to $+\infty$ and the other factor to$~0$), there is absolutely no ambiguity about the sum of an infinite number of terms, all (exactly) equal to$~0$; this sum is$~0$, always*. The value of an infinite sum is defined as the limit (if it exists) of the sequence of finite partial sums of terms. Since in the case under consideration all those partial sums are$~0$, their limit (and therefore the infinite sum) is clearly$~0$.

Infinite sums of equal terms are not the same thing as multiplying that value by$~\infty$.

By the way there is no ambiguity either about a sum with $0$ terms, even if that term would potentially be $+\infty$; since the term is never actually produced, its potential problem never occurs, and the empty sum is$~0$. One could think of a sum like $\sum_{n=1}^0\frac1{n-1}$ where the term for $n=1$ would be problematic, yet the summation unambiguously has value$~0$. Similarly products with no factors at all are$~1$, as in $0!=1$. For some reason that I don't want to get into here some people object to the this evaluation of the empty product if it takes the form $0^0=1$, even though it is the exact same product as $0!$ (namely one without any factors at all).

*Technically excluding some situations where the limit manages to be not uniquely defined, which could happen if it is taken in a space with non-Hausdorff topology. You can ignore this.