# 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

I'd love if I can have this confusion sorted out. Thanks!

• I haven't watched the video, but note that $\mathrm{lim}_{n \rightarrow \infty} (0a_n) = 0$ Jun 12 '14 at 8:55
• Indeterminate form is better thought as just a way to remember when limit do not distribute over algebraic operation. In this particle case, the only arithmetic operation is multiply, which obviously distribute: $\sum\limits_{n=1}^{\infty}a_{n}x^{n}=\lim\limits_{k=1}^{\infty}\sum\limits_{n=1}^{k}a_{n}x^{n}$ $=\lim\limits_{k=1}^{\infty}(x\sum\limits_{n=1}^{k}a_{n}x^{n-1})=(\lim\limits_{k=1}^{\infty}x)(\lim\limits_{k=1}^{\infty}\sum\limits_{n=1}^{k}a_{n}x^{n-1})$ which result in $0\times 0$ so not an indeterminate form at all.
– Gina
Jun 12 '14 at 9:25
• Hang on, if $x=0$ we can't even do the ratio test to begin with. Jun 12 '14 at 12:18
• There is a difference between $0$ and something whose limit is $0$. The "indeterminate forms" are about the latter case, not the former. Jun 12 '14 at 14:55
• I want to point out one more thing that may help: The limit here is not exactly the same thing as the limit that one encounters in a calculus class. In this video, she is talking about the limit of a sequence, not the limit of a function. Sequences themselves do not necessarily have an "indeterminate form". In the case of sequences, what you have to do to find a limit is find a number that the sequence gets close to at infinity (not at a point, as is common in Calculus). For example, the sequence 1, 1/2, 1/3, ... gets close to 0. So does 0, 0, 0 ... Jun 12 '14 at 14:57

$\sum_{n=1}^{\infty}a_{n}$ 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 $\lim_{n\rightarrow\infty}s_{n}=0$

• +1,'...Well done, good and faithful servant!...' (Matt. 25:21) Jun 12 '14 at 10:37
• While this explains the convergence, it doesn't really answer the OP's question, which was how the ratio test works in this case. The correct answer seems to be this one Jun 12 '14 at 18:31
• @boywholived Thank you. I allready enjoy it very, very much if human beings say things like that to me. It makes me willing to be a servant of anyone! Now try to imagine how much I will enjoy it if this is said to me by the One I have in mind right now. Jun 12 '14 at 19:44
• It's not just "formally the limit", it's by definition the limit. Jun 13 '14 at 7:47

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.

• +1 "Infinite sums of equal terms are not the same thing as multiplying that value by ∞." Jun 12 '14 at 16:53
• There's also the concept of a "strong" zero which, when multiplied by anything (even something undefined or meaningless) gives zero. e.g. Concrete Mathematics considers the Iverson bracket to be strongly zero when it's zero. But this can be interpreted as a notational convention for "don't include the terms where this thing is zero" rather than actually meaning to multiply the thing by zero.
– user14972
Jun 12 '14 at 18:00
• Incidentally, I am one of the sorts who object to $0^0=1$... but since you've explicitly stated your notation is referring to repeated products rather than any of the other senses one can mean by exponentation, I would agree that $0^0=1$.
– user14972
Jun 12 '14 at 18:01

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 $0$.

• so are you saying that if x=0, then the power series will all turn into zeros so 0⋅∞ = 0 is justified in this particular case? Jun 12 '14 at 9:06
• Yes, in this particular case it is. It is not productive to think of this limit as a product of limits; the expression $0\cdot\infty$ has no well-defined meaning. Try not to treat the symbol '$\infty$' as if it were a number, this causes a great deal of confusion. Jun 12 '14 at 9:08
• @M.Samuel When you say that $0\cdot \infty$ is unknown, what you really mean is that if $a_n$ is a sequence with $\lim_{n\to\infty} a_n = 0$ and $b_n$ is a sequence with $\lim_{n\to\infty} b_n = \infty$, then $lim_{n\to\infty} a_n b_n$ is unknown without knowing how $a_n$ and $b_n$ behave compared to each other. If the sequence $a_n$ is zero for all $n$, then the limit would be simplified to $\lim_{n\to\infty} 0b_n = \lim_{n\to\infty} 0 = 0$. The "$0\cdot\infty$ is unknown" is only about when the "$0$" is a sequence converging to zero, not when the "$0$" is an actual constant $0$. Jun 12 '14 at 9:23

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 $n!0^n=n!\times0=0$

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.

0 to any power is 0 therefore the sum of zeros is zero

• This is incorrect. Zero raised to a negative number is undefined; for zero raised to the zeroth power, see this. Jun 13 '14 at 5:09