Im trying to show that the series
sum(n=1_infinity) 1/(4n^(1/3) -1) > sum(n=1_infinity) 1/n term by term
As part of the proof of divergence by direct comparison
so I need to show that
n > 4n^(1/3) -1 in order for 1/n < 1/(4n^(1/3) -1)
so i will compare the rates of change of the denominators ( both series are decreasing on the interval [1,inf.))
n > 4n^(1/3) -1
by taking the first derivative of both sides
1 > 4/(3*n^(2/3))
take the limit of both sides as n--> infinity
1 > 0
so there exists some value N[1, inf.) that, n >/= 4n^(1/3) -1
which would show that for value greater than N[1,inf) 1/n </= 1/ (4n^(1/3) -1)
or so I think....😕
Originally posted by joe shmoIf you rearrange the terms you get that what you want to prove is that (n+1)/4 > n^(1/3) which is equivalent to
Im trying to show that the series
sum(n=1_infinity) 1/(4n^(1/3) -1) > sum(n=1_infinity) 1/n term by term
As part of the proof of divergence by direct comparison
so I need to show that
n > 4n^(1/3) -1 in order for 1/n < 1/(4n^(1/3) -1)
so i will compare the rates of change of the denominators ( both series are decreasing on the interval [1,i ...[text shortened]... would show that for value greater than N[1,inf) 1/n </= 1/ (4n^(1/3) -1)
or so I think....😕
[; \left( \frac{n+1}{4}\right)^3 > n \Leftrightarrow \frac{(n+1)^3}{64} > n \Leftrightarrow \frac{n^3+3n^2+3n+1}{64} > n \Leftrightarrow \frac{n^3+3n^2-61n}{64} > 0 \Leftrightarrow ;]
[; \Leftrightarrow n^3+3n^2-61n > 0 \Leftrightarrow n(n^2+3n-61) > 0 \Leftrightarrow n^2+3n-61 > 0 ;]
But I think that this quadratic has real zeros and so either what you are trying to prove is wrong or I made a mistake... 😕
Use this in order to see my formulas: http://thewe.net/tex/
Originally posted by adam warlockhmmm...well I didn't download the software
If you rearrange the terms you get that what you want to prove is that (n+1)/4 > n^(1/3) which is equivalent to
[; \left( \frac{n+1}{4}\right)^3 > n \Leftrightarrow \frac{(n+1)^3}{64} > n \Leftrightarrow \frac{n^3+3n^2+3n+1}{64} > n \Leftrightarrow \frac{n^3+3n^2-61n}{64} > 0 \Leftrightarrow ;]
[; \Leftrightarrow n^3+3n^2-61n > 0 \Leftrightarrow ...[text shortened]... made a mistake... 😕
Use this in order to see my formulas: http://thewe.net/tex/
but the polynomial n^3 +3n^2 -61n +1 = 0
has 3 real solutions
And there is only one in the domain that i am worried about ( it occurs between n= 6 & n=7 if i remember correctly
so which ever series is above will continue to be above for the rest of the domain in question....or at least I hope..
Okay, returning to your original idea, it is quite simple to show that for n>8 it is fulfilled n>4n^1/3 -1, or n-4n^3+1>0. Let's substitute n^1/3 = x. Then we have x^3 -4x>-1. Since x is positive and greater or equal to one (n is greater than 1) we have (x-2)(x+2)x>-1. The LHS is always positive for x>2 (n>8).
[for notation purposes i will use G(x) to replace "sum of all x as x begins at 1 and increases by one, tending towards infinity" i.e. it is the normal capital sigma we use with n=1 below and infinity on top.]
you are looking to show that G(1/[4n^(1/3)-1]) diverges. so we wish to compare it term by term to G(1/n).
first, however, consider G(1/4n)... if you write it out you will see quickly that G(1/4n) = 1/4 * G(1/n), which clearly diverges.
but then term by term comparison is much easier... for all n greater than or equal to 1, 4n^(1/3) < 4n. furthermore, [4n^(1/3) - 1] < 4n^(1/3) < 4n.
so then, every denominator in our series is strictly smaller than that of a divergent series, and every numerator is the same as that divergent series. so our series is term-by-term larger than a divergent series, and so also must diverge.
note that in dealing with comparisons to the harmonic series, one of the most useful things to note is that a constant multiplier that does not increase as n increases is inconsequential to the divergence or convergence of the sequence. and so for a term-by-term comparison test it's often useful to multiply the harmonic series by a useful constant of our choice.
cheers!
Originally posted by Aetheraelok..will ponder in light of information...but for now cheers as well!!!
[for notation purposes i will use [b]G(x) to replace "sum of all x as x begins at 1 and increases by one, tending towards infinity" i.e. it is the normal capital sigma we use with n=1 below and infinity on top.]
you are looking to show that G(1/[4n^(1/3)-1]) diverges. so we wish to compare it term by term to G(1/n).
...[text shortened]... s often useful to multiply the harmonic series by a useful constant of our choice.
cheers![/b]