Stumped by a Subscriber's Question, Convergence of Series with Comparison Test?

In my QnA post, I was asked a question by a subscriber Nelson Talukder that stumped me. It also demonstrates an important lesson - don't put /too/ much faith in Wolfram|Alpha.

Big thanks to Math.SE posters Hans Lundmark and kccu for helping me with this problem.

Music:
Blue Wednesday - Cereal Killa

Music: K. Lumsy
Composer: Grant Kirkhope
Platform: Nintendo 64

Outro:
"Lateralus" as performed by Sakis Strigas
Originally by Tool

[nelsontalukder]

Thank you very much for this! It is nice to know I was not the only one perplexed by this. I was beaming the entire way through, thank you for choosing my question! Brilliant video!^^

[EssentialsOfMath]

This is one of those instances where, the more advanced you are, the more basic stuff you forget.

[EpicMathTime]

At 3:46 I meant one over the thing I said. (1/x^(1.0001)).

Lemmy is God.

[nickmattson]

"Recently I axed you..."


I love America, the land of ruffs and axes

[shohamsen]

First simplify the inequality to get x < ( ln x)^{1000} .Then plug in x=e^y, this gives us e^y < y^1000. Note that since y is the variable, the question is equivalent to asking which grows faster exponential or polynomial.

[Eva-ezks]

It's interesting, Wolfram Alpha computes both intersections for xlnx and x^1.0001, but only the first intersection of their reciprocals.

[Quasarbooster]

Very good lesson indeed! Big numbers and 'puters don't mix.
When you mentioned that the bottom inequality may not be true I searched for the points of intersection as well. After some steps you'd get x^.0001 = ln (x) and there's really no good way to solve unless you know about the Lambert W function. If you do, then you can solve to get x = e^(W(-.0001)/-.0001). Since I knew what the graph of W(x) looks like I knew that W(-.0001) would give one output at almost -.0001 and another output further down (it was around -11). The second output explains why Wolfram didn't notice the second pt of intersection at around e^110000

[DistortedV]

and ladies and gentleman, this is why we need theory over experiment

[Anokosciant]

my god this serie diverges so slowly jesus

[perappelgren]

Great video! Fruitful to learn that even professionals sometimes run into optical illusions :-)

The use of the Lambert W function has already been commented upon, so this post is not unique wrt W, but anywho...

Basically, to be able to say something about the inequality, we want to solve the equation

(1) 1/(xln(x)) = 1/x^1.0001 for x >= 3

We readily see that ln(x) and x^1.0001 are both > 0 whenever x >= 3. So by multiplying both membra by x, we get

(2) 1/ln(x) = 1/x^0.0001

Setting ε = 0.0001 and cross-multiplying we now have

(3) x^ε = ln(x)

Now, letting x = exp(u) transforms (3) into

(4) exp(u*ε) = u

Diving (4) by exp(u*ε) yields (5) and after some housekeeping we arrive at (5.1)

(5) 1 = u*exp(-ε*u)
(5.1) -ε = -ε*u*exp(-ε*u)

Now, by the notion of the Lambert W function, we know that whenever y = x*exp(x), x is given by x = W(y) (for "reasonable" real values of y). The right hand side contains -ε*u in lieu of x, while -ε plays the role of y. Hence, from (5.1) it follows that

(6) -ε*u = W(-ε)
(6.1) u = -W(-ε)/ε

Substituting back into x (see (4)) gives us the "solutions", namely

(7) x = exp(-W(-ε)/ε)

A note of caution is needed here. The Lambert W function is two-branched; for all (real) x-values in [-1/e 0), there are two possible values of W, corresponding to the two portions of the curve y = x*exp(x) on each side of the minimum in x = -1. As ε = 0.0001, and as the argument of W in this case equals -ε, we clearly find ourselves in the two-branch range.

[gergodenes]

I know, old video, but here's what I came up with (probably similar to the solution on stack exchange, it's pretty straightforward):
Let's examine the inequality, and by doing some algebra, we come up with the inequality
ln(x)^10000 >= x (assumed that neither is 0 or negative)
from this follows x/ln(x)^10000 <= 1.
If we take the limit of x/ln(x)^10000, as x approaches infinity, with 10000 iterations of L'Hospital's rule, we get the limit of x/10000! as x approaches infinity. From this, it's obvious that the limit goes to infinity, meaning that the inequality does not hold for large enough x.
QED? :)

[BenjaminKuruga]

Calling 4.3x10^50669 a large number is an ... understatement, but I guess it still didn't over-run a googolplex :P

[alkankondo]

Great video! Also, I like how you use selections from the DK64 soundtrack for your background music. It's one of my favorite games on the Nintendo 64!

[proghostbusters]

Just realized the outro was Lateralus! I see you're a man of culture.

[crashplague]

EMT doesn't have a jawline, he has a jawplane.

[swozzlesticks]

Whenever I see those really tiny numbers (1.0001) in an exponent I immediately get suspicious that it's just trying to fool computation. It's not always the case but it's helped me a bunch.

[MrRyanroberson]

Even before wolfram my first thing is to note that because therange is positive we can just flip the inequality and the fractions to get x^1.0001 <= x ln(x), divide by x, and it's clearly eventually false because growing polynomials grow faster than logarithms universally

[coolcat]

this is old, but I have another (maybe informal) way of looking at this:
Suppose the inequality holds. This means we have x^a <= ln(x) where a = 0.0001 and x >= 3 (after some algebra). We can then say x^a / ln(x) <= 1. However, by taking the limit of x^a / ln(x) as x -> infinity, we can verify that the limit ends up approaching infinity by using l'Hopital's Rule. This contradicts x^a / ln(x) <= 1, so the inequality shouldn't hold.

[ThAlEdison]

Setting the derivatives equal to each other, you can see that the graphs are parallel at 10^40, 000 from there you can sort of work things out
ln(10^40000)~=92, 103
And (10^40000)^(0.0001)=10, 000
Playing it forward, 116, 672^10, 000 is past the point where the two graphs cross
ln(116672^10, 000)~=116, 671
And (116672^10000)^0.0001=116, 672

[TVbr]

rule : for every a>0 lnx/x^a tends to 0 as x tend to infinity, thus after some number x0, x^a>lnx