Why is infinity multiplied by zero considered zero here? I watched an online video lecture by some professor and she was solving a convergence problem of the power series $$\sum_{n=1}^\infty n!x^n,$$  i.e., she was finding the values of $x$ for which this power series is convergent.
She did the ratio test and winded up with $(n+1)x$ and now she started to compute the limit of this thing as $n$ approaches infinity and that's where my confusion started!
She said that :
i) If $x \neq 0$, the limit is infinity (I agree with that).
ii) If $x = 0$, the limit is $0$ (this is what I don't agree with because if $x = 0$, and $n$ approaches infinity, I should have the indeterminate form of $0\cdot\infty$. So why did she decide to make it zero?
P.S. Here is the video I'm talking about and this problem starts approximately after 6 min
https://www.youtube.com/watch?v=M8cojIKoxJg
I'd love if I can have this confusion sorted out. Thanks!
 A: If $x\neq0$ then the sequence $(n+1)x$ increases without bound as $n$ increases, so the sequence tends to infinity.
If $x=0$ then $(n+1)x=0$ for all $n$, so the sequence is constantly $0$. Hence its limit is also $0$.
A: $\sum_{n=1}^{\infty}a_{n}$ 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 $\lim_{n\rightarrow\infty}s_{n}=0$
A: The sum $\sum_{n=1}^\infty n!x^n$, is equal to zero at $x=0$ because if $x=0$ then the sum simplifies to $\sum_{n=1}^\infty 0=0$
To show that it simplifies to this:
$n!x^n$ for $x=0$ becomes $n!0^n$. Since $0^n=0$ for all $n\ne0$ and that the sum starts from $1$ meaning that $n$ is never $0$. This means that $n!0^n=n!\times0=0$
A: 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.
A: Intuitively:


*

*When x=0 and n=1, the term is 0.

*When x=0 and n=2, the term is 0. 0 + 0 = 0.

*When x=0 and n=3, the term is 0. 0 + 0 = 0.


Keep increasing n forever, keep adding zeros, will the value of the sum ever not be 0?
Therefore the limit of this sequence is 0.
A: 0 to any power is 0 therefore the sum of zeros is zero
