What does equal?
You could rewrite the equation as
This is then written as .
Which results in !!!.
Note by
Sharky Kesa
5 years, 5 months ago
No vote yet
1 vote
</code>...<code>."> Easy Math Editor
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
Sort by:
Top NewestThis is an infinite geometric series which has ratio r=2>1. So this doesn't converge, and as the result we get an non-sense answer.
Log in to reply
I'm sorry but you can't just use the distributive laws when dealing with infinitely many terms. A simple proof that this is wrong is that a sum diverges when its partial sums diverges. Clearly the partial sums are 2n−1 which approaches to infinity as n does, therefore the sum does not exist and is undefined.
Or instead you are talking about something much more complicated like in a different number system, like 2-adic numbers. :P
Log in to reply
Damn, we communicated before Schol of Excellence???
Log in to reply
I do think this answer is a bit weird, but the method seems correct, since it is an infinite series! and does not end.
Log in to reply
Just an illusion.....,
Confine up to 5 or 6 terms and try to evaluate..
InitiallyTrolled me too..,
Log in to reply
I had a similar derivation which I am not so sure about. My first note on Summations
Log in to reply
these laws are not valid when we are dealing with infinite sums.
Log in to reply
ahh I like these fallacious proofs - they really test your observation. I think the reason why you got -1 was because the factorization only works if the expression is finite.
Log in to reply
It has to do with the fact that you're ignoring the last term, which is infinitely large and therefore very important
Log in to reply
There is no largest term.
Log in to reply
I do not see where −1 came from.
EDIT: For a proof of where this summation went wrong, check my comment down here.
Log in to reply
Remove the brackets and you get 2+4+8+16+32+...-1-2-4-8-16-32. The +2 and -2 cancel each other out, so does +4 and -4, +8 and -8, etc. until all that's left is -1
Log in to reply
This is obviously false, we still have a term in the first sequence missing.
If the RHS ends, let's say, in 2n, the LHS will end in 2n+1. This leads us that the Σ=2n+1−1, which is true.
But we cannot evaluate a limit for n→∞, because Σ would tend also to infinity.
Log in to reply
Since it was (2−1)(1+2+4+8+16+32+...), you just simplify it. 2(1+2+4+8+...)−1(1+2+4+8+16+...) so it is 2+4+8+16+32...−1−2−4−8−16.... 2 and −2, 4 and −4, etc. all cancel each other out until −1 remains
Log in to reply
However, the proof for 1+2+3+4+5+6+…∞=12−1, requires the assumption that the R.H.S. does not end. The point of infinite sequences is that they do not end. Even the proof for 9.99999…=10 requires the 9s to never end.
Log in to reply
Nice logic!!
Log in to reply