$\infty / \infty$

Calculus Level 2

$\large \dfrac{1+1+1+\cdots}{2+2+2+\cdots}=\, ?$

Clarification : Both the numerator and denominator are infinite sequences.

0.5 1 2 Undefined

5 solutions

Rakibul Raihan
Jan 23, 2016

infinite/infinite=undefined

Dude...infinite/infinite is indeterminate, not undefined!

Abhishek Kumar - 5 years, 4 months ago

Nooo, look. Let's define $S^{1}_{n}=\Sigma_{i=1}^{n}1$ and $S^{2}_{n}=\Sigma_{i=1}^{n}2$ . Both are well defined sequences that satisfy the relation: $S^{2}_{n}=2S^{1}_{n},$ for all $n\in N;$ therefore, if we define $S_{3}^{n}=\dfrac{S^{1}_{n}}{S^{2}_{n}}=\dfrac{1}{2}$ for all $n\in N.$ . Since $S^{3}_{n}$ is independent of $n$ , then we have the limit: $\lim_{n\to\infty}S^{3}_{n}=\dfrac{1}{2}.$

You're making a huge mistake, which is equivalent to affirm that $\lim_{x\to x_{0}}\dfrac{f(x)}{g(x)}=\dfrac{f(x_{0})}{g(x_{0})},$ when both $f(x),\;g(x)$ are discontinuous at $x=x_{0}.$

Ernesto López Fune - 5 years, 4 months ago

Your definition of $S_n^3$ implies that you're taking the number of terms in the numerator and denominator to be equal; that is, you're computing the limit of $\frac{1}{2}, \frac{1+1}{2+2}, \frac{1+1+1}{2+2+2}, \ldots$ . This is indeed $\frac{1}{2}$ , but that's not what the problem is asking. The numbers of terms aren't necessarily going in the same rate. It's possible that the expression is to be interpreted as $\frac{1+1}{2}, \frac{1+1+1+1}{2+2}, \frac{1+1+1+1+1+1}{2+2+2}, \ldots$ , for example, with a different limit (1), so it is undefined.

This is the same issue with $\displaystyle\int_{-\infty}^{\infty} x \, dx$ . Some people argue that this means $\displaystyle \lim_{n \to \infty} \int_{-n}^n x \, dx = \lim_{n \to \infty} 0 = 0$ , but this is incorrect, since the two boundaries can approach infinity in different rates; for example, it's possible to interpret the expression as $\displaystyle \lim_{n \to \infty} \int_{-n}^{2n} x \, dx = \lim_{n \to \infty} \frac{3}{2} n^2 = \infty$ , so it's also undefined.

Ivan Koswara - 5 years, 4 months ago

yes, i made a mistake assuming the same number of element in every sum. i apologize...

Ernesto López Fune - 5 years, 4 months ago

I want to know after doing both summations,putting limits where is the discontinuity and why we should not apply limits

DIVYA ASHISH - 5 years, 4 months ago

@DIVYA ASHISH consider this: $\lim_{n\to\infty} (n-n)=0$ but consider this: $\lim_{(x,y)\to\infty}(x-y)$ This is not≠0, similarly, here there are two different series so there is not one variable in the limit but two.

Aareyan Manzoor - 5 years, 4 months ago

Suppose $n,m$ the number of elements of each series: $S^{1}_{n}=\sum_{i=1}^{n}1=n$ and $S^{2}_{n}=\sum_{i=1}^{m}2=2m.$ Defining $S^{3}_{n,m}=\dfrac{S^{1}_{n}}{S^{2}_{m}}=\dfrac{1}{2}\dfrac{n}{m}.$

You can see that the limit $\lim_{(n,m)\to\infty}S^{3}_{n,m}$ does not exists. It depends on the "trajectory". For example: $n=m\Rightarrow \lim_{(n,m)\to\infty}S^{3}_{n,m}=\dfrac{1}{2}.$

$n=m^{2}\Rightarrow \lim_{(n,m)\to\infty}S^{3}_{n,m}=\infty.$

$m=n^{2}\Rightarrow \lim_{(n,m)\to\infty}S^{3}_{n,m}=0.$

Ernesto López Fune - 5 years, 4 months ago

Hi I am confused on your usage of S notation, is this an arithmetic sequence? Forgive my ignorance..

Marcus Chan - 3 years, 3 months ago

You made a fundamental mistake from the beginning when you claimed that $S^{2}_{n} = 2S^{1}_{n}$ . Their relationship is not that.

Evan Huynh - 5 years, 4 months ago

That equation is correct for all $n$ . Note that the definitions there are finite sums and thus defined.

Ivan Koswara - 5 years, 4 months ago

∞/∞ is actually 1.

Windows 96 - 1 year, 8 months ago

No, $\frac{\infty}{\infty}=\text{indeterminate form}$ .

Ved Pradhan - 11 months, 4 weeks ago

relevent wiki

Aareyan Manzoor - 5 years, 4 months ago

Rather indeterminate

Deepak Kumar - 5 years, 4 months ago

Well , Both are constant GP's , so by sum of infinite GP , it comes out to be $\frac{1}{0}$ , which is surely indeterminate .

A Former Brilliant Member - 5 years, 4 months ago

individually numerator and denominator is infinite but what about their ratio?

Abhishek Kumar - 5 years, 4 months ago

Pranav Jayaprakasan UT
Jan 30, 2017

Using infinite geometric series, a/(1-r) We get that 1+1+1+1......= 1÷0. Hence it is undefined

You cant use infinite sequence for this because r must be a fraction i.e (-1,1)

Swapnil Vatsal - 4 years ago

Thebostinteaparty .
Feb 4, 2020

1+1+1+1... = inf while 2+2+2+2... = inf. but inf/inf = N/a

Juan Manuel Fabregat Moreno
Jan 28, 2016

for being 1/2 we´ll have to take the same terms in both sides of the operation

Can you elaborate, please?

Konstantin Zeis - 4 years, 3 months ago

I agree this is a bad answer. What about arguing the sequences both add up to infinity(they are both infinite sequences) However, there are different types of infinity: in this example, infinity=2xinfinity. Moreover, it is not a number but rather a concept, used to represent the biggest number(there is no biggest number). It is used as a number sometimes and is helpful to mathematicians, but it holds many different 'values' and can not be used in arithmetic as it is not a true number. This problem is relevant to the Wiki on infinity divided by infinity. Is this more helpful?

Zoe Codrington - 2 years, 7 months ago

. .
Mar 22, 2021

$\displaystyle \frac { 1 + 1 + 1 + 1 + \cdots } { 2 + 2 + 2 + 2 + \cdots } = \frac { \infty } { \infty } \ne 0.5, 1, 2$ .

$\displaystyle \frac { \infty } { \infty } \text { is undefined }$ .

∞ / ∞ \infty / \infty ∞ / ∞

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