# would this be sufficient

joe shmo
Posers and Puzzles 02 Apr '09 18:46
1. joe shmo
Strange Egg
02 Apr '09 18:46
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....ðŸ˜•
Baby Gauss
02 Apr '09 23:238 edits
Originally posted by joe shmo
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....ðŸ˜•
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 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... ðŸ˜•

Use this in order to see my formulas: http://thewe.net/tex/
3. joe shmo
Strange Egg
03 Apr '09 00:16
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... ðŸ˜•

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..
4. 03 Apr '09 07:39
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).
5. Palynka
Upward Spiral
03 Apr '09 09:58
First show that:
1/(4n^(1/3)-1) > 1/n, for all n>6

Then highlight the fact that H_inf - H_6 = inf (note H_n:harmonic number)

So sum(n=7_infinity) 1/n < sum(n=7_infinity) 1/(4n^(1/3) -1) = inf
=> sum(n=1_infinity) 1/(4n^(1/3) -1) = inf
6. 05 Apr '09 03:332 edits
[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!
7. joe shmo
Strange Egg
05 Apr '09 06:53
Originally posted by Aetherael
[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]
ok..will ponder in light of information...but for now cheers as well!!!