Taylor expansion isn't right??

Hello everyone!! I had an extremely weird question......pls help

We all know that sum of rational numbers is rational......we also know that e is irrational.....

But, e = 1 + 1 + 1/2 + 1/6 + 1/24 + 1/120 + 1/720 + 1/5040 + 1/40320 .......... The RHS is consisting of all rational numbers but e is irrational......How is it possible??? Pls help!! We can extend this to any Taylor Series........eg. Sin(pi/6) is rational but it's taylor series expansion consists Pi in all terms and so the expansion is irrational contradicting that Sin(pi/6) is rational......

What to do??

#Calculus

Easy Math Editor

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Comments

I recently made a Challenge Master note regarding this.

An operation / property has closure if performance of the operation always returns an element of the set. For example,

the set of positive integers is closed under finite addition.

the set of positive integers is not closed under infinite addition.

the set of open sets is closed under finite intersection.

the set of open sets is not closed under infinite intersection.

In this case, the set of rational numbers is closed under finite addition, but not closed under infinite addition (even with the assumption that the sum is finite). As an example, $\pi = 3 + 0.1 + 0.04 + 0.001 + 0.0005 \ldots$ .

Calvin Lin Staff - 3 years, 3 months ago

@Calvin Lin Ok Sir, thanks for the information!!! But, it naturally arises a few questions........

1.) Is this is a known theorem?? If so, what is the proof?? If not, is it just a convention made in mathematics to make things easier??
2.) If we now know that set of rational numbers is not closed under infinite addition, that means summation of infinite rational numbers can along with irrationals, also lead to a purely imaginary number...??
3.) In the example of pi = 3 +0.1 +0.04 + 0.001 ....... We know that pi is the limit of the sequence a_n = sum of the place value of first n digits of pi............ But, we ourselves do not know every digit of pi...........It is kind of like , instead of approximating, by putting an equal to sign, we are saying we know each digit of pi.....And we also know that a limit existing does not specify that the function actually exists there.....Also, how do we know that the sequence converges......?? What sort of convergence test can lead us there??

Aaghaz Mahajan - 3 years, 3 months ago

@Calvin Lin Sir, please do reply...

Aaghaz Mahajan - 3 years, 1 month ago

@Aaghaz Mahajan – 1) You can consider it a theorem. It's essentially just a proof that "the sum of infinitely many rational numbers need not be rational".

2) No. the sum of infinitely many real numbers is always real. The real numbers are closed under (infinite) addition.

3) I have no idea what you're getting at here.
If your concern is that we don't know the exact digits of pi, then take any irrational number that we know the digits of, like $\sum 10^{ - n^ 2 }$ .
We know the sequence converges because it is a strictly increasing sequence that is bounded above. Alternatively, apply the ratio test - The ratio of (non-zero) successive terms is at most 0.9.

Calvin Lin Staff - 3 years, 1 month ago

@Calvin Lin @Pi Han Goh @Chew-Seong Cheong @Otto Bretscher @Sandeep Bhardwaj Pls Help!!!

Aaghaz Mahajan - 3 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$