Since the sum converges absolutely, we’re allowed to rearrange the terms: 2+1/4+1/2+1/16+1/8…=2+1/2+1/4+1/8+1/16… Since 1+1/2+1/4+1/8+1/6…=2 by geometric series(1+x+x^2…=1/1-x for |x|
Like the other guy said, it doesnt alternate so if it converges it should converge absolutely. But I think I solved it differently. So if we take the general term, but seperate them into the even terms and the term right after. We can simplify them to these. a_(2n) = 2^(-2n + (-1)^(2n)) a_(2n) = 2^(-2n+1) a_(2n+1) = 2^(-2n-1 + (-1)^(2n+1)) a_(2n+1) = 2^(-2n-1-1) a_(2n+1) = 2^(-2n-2) If we add up the even terms (even value of n) and the (odd) term that is right after it, we can get the general term b_n that would be an equivolent series which can be simplified as b_n = 2^(-2n+1) + 2^(-2n-2) b_n = (2 + 2^-2)*2^(-2n) b_n = (2 + 1/4)*4^(-n) b_n = 9/4 * (1/4)^n After that it should be clear that the equivolent series is geometric and it should be fairly easy that the series adds up to 3 when summing from 0 to infinity if you plug it into the equation for sum of a geometric series.
We can seen here Stolz Cesaro theorem, if I take cases where a_n not negative. lim (a_n)^1/n= lim exp(log a_n/n). If exists finite lim ( log a_n+1-log a_n)/ (n+1-n) , then allways exists lim log(a_n)/n and limits are the same. So if exists finite limit in root test, then always exist limit in ratio test, and this limit is the same. But there is only implication not ekvivalention. But if exists finite limit in root test, the limit in ratio test may not always exist. I never read about it in literature, but I think, if limit in root test is 1, then limit in ratio test is always 1 or doesn't exist. If limit in ratio test is 1, then limit in root test will be 1,(if root test is good for useing, a_n not negative) .....,
LMAO 1:30 this is very not stonks. I was crying from laughter. I love this guy so much
Personally I used comparison test when comparing this to the geometric series (a multiple of 1+1/2+1/4+...)
You are great sir,
Can you give an example of such as series?
Do you mean by that that these can not be computed or the convergence is slow?
This seems to converge to 3. Do you have a proof for this?
Since the sum converges absolutely, we’re allowed to rearrange the terms:
2+1/4+1/2+1/16+1/8…=2+1/2+1/4+1/8+1/16…
Since 1+1/2+1/4+1/8+1/6…=2 by geometric series(1+x+x^2…=1/1-x for |x|
Like the other guy said, it doesnt alternate so if it converges it should converge absolutely. But I think I solved it differently.
So if we take the general term, but seperate them into the even terms and the term right after. We can simplify them to these.
a_(2n) = 2^(-2n + (-1)^(2n))
a_(2n) = 2^(-2n+1)
a_(2n+1) = 2^(-2n-1 + (-1)^(2n+1))
a_(2n+1) = 2^(-2n-1-1)
a_(2n+1) = 2^(-2n-2)
If we add up the even terms (even value of n) and the (odd) term that is right after it, we can get the general term b_n that would be an equivolent series which can be simplified as
b_n = 2^(-2n+1) + 2^(-2n-2)
b_n = (2 + 2^-2)*2^(-2n)
b_n = (2 + 1/4)*4^(-n)
b_n = 9/4 * (1/4)^n
After that it should be clear that the equivolent series is geometric and it should be fairly easy that the series adds up to 3 when summing from 0 to infinity if you plug it into the equation for sum of a geometric series.
Does these series converge to a transcendental number? Why don't you even try to answer this question for any single convergent series?
It’s not the point, many convergent series are impossible to evaluate
@@drpeyam Can you give me an example?
@@Guille-g8l sum(n/(3^n+1))
who wings
We can seen here Stolz Cesaro theorem, if I take cases where a_n not negative. lim (a_n)^1/n= lim exp(log a_n/n). If exists finite lim ( log a_n+1-log a_n)/ (n+1-n) , then allways exists lim log(a_n)/n and limits are the same. So if exists finite limit in root test, then always exist limit in ratio test, and this limit is the same. But there is only implication not ekvivalention. But if exists finite limit in root test, the limit in ratio test may not always exist.
I never read about it in literature, but I think, if limit in root test is 1, then limit in ratio test is always 1 or doesn't exist. If limit in ratio test is 1, then limit in root test will be 1,(if root test is good for useing, a_n not negative) .....,
you can also use for convergention geometric series as majorant
Delicious breakfast early evening limsup
First