Should I post or Wait?

nasi112

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I'm asking because it's a bit odd that I'm the only one posting. Should I hold off until the site is fixed? If that's the case, I'll stop with this question.

n=2(n!)2(2n)!x2n+1\sum_{n=2}^{\infty}\frac{(n!)^2}{(2n)!}x^{2n+1}
I'm having some trouble solving this sum, particularly with the absolute value. Let me show you what I mean.

If I use the ratio test I get limn((n+1)!)2(2(n+1))!x2(n+1)+1(2n)!(n!)2x2n+1\displaystyle \lim_{n\to \infty}\left|\frac{((n + 1)!)^2}{(2(n + 1))!}x^{2(n + 1)+1}\frac{(2n)!}{(n!)^2x^{2n+1}}\right|

I'll skip some simplification.

limn(n+1)2(2n+2)(2n+1)x2=14x2\lim_{n\to \infty}\left|\frac{(n + 1)^2}{(2n+2)(2n+1)}x^2\right| = \left|\frac{1}{4}x^2\right|
The ratio test states that this converges only when 14x2<1\left|\frac{1}{4}x^2\right| < 1

If I apply the absolute value rule along with the inequality, it leads to strange results.

1<14x2<1-1 < \frac{1}{4}x^2 < 1
I don't feel this inequality is correct because x^2 can't be negative.
 
I'm asking because it's a bit odd that I'm the only one posting. Should I hold off until the site is fixed? If that's the case, I'll stop with this question.
Keep asking. It keeps the site going, otherwise none of the helpers would bother checking it.
The ratio test states that this converges only when 14x2<1\left|\frac{1}{4}x^2\right| < 1

If I apply the absolute value rule along with the inequality, it leads to strange results.
1<14x2<1-1 < \frac{1}{4}x^2 < 1I don't feel this inequality is correct because x^2 can't be negative.
That doesn't matter; it just means that 1<14x2<1-1 < \frac{1}{4}x^2 < 1 and 014x2<10\le \frac{1}{4}x^2 < 1 are equivalent.

That is, the set of x for which the graph is in this region,
1750551861890.png
is the same as the set of x for which the graph is in this region,
1750551897124.png
because it is never in this region:
1750551975756.png
So both the blue and the green are "correct" ways to say the same thing. The latter is just simplified.

The big question is, what is the solution (for x)?
 
I'll restate the inequality 14x2<1|\frac{1}{4}x^2| < 1

If this can be simplified to 1<14x2<1-1 < \frac{1}{4}x^2 < 1, then I would typically choose any value within that interval. For example, I might pick 0.9=14x2-0.9 = \frac{1}{4}x^2, but I believe that approach is incorrect. That's just the method I'm used to applying when dealing with inequalities.

I understand the graphs you shared, but I’d like to take a closer look at what I’m trying to analyze.

When I examine the role of the absolute value in the inequality 14x2<1|\frac{1}{4}x^2| < 1 I realize it's not necessary. Is it valid to simply remove it and continue solving from there?

14x2<1\frac{1}{4}x^2 < 1
x2<4x^2 < 4
x<2|x| < 2
2<x<2-2 < x < 2
I edited the -0.9.
 
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I'll restate the inequality 14x2<1|\frac{1}{4}x^2| < 1

If this can be simplified to 1<14x2<1-1 < \frac{1}{4}x^2 < 1, then I would typically choose any value within that interval. For example, I might pick 0.9<14x2-0.9 < \frac{1}{4}x^2, but I believe that approach is incorrect. That's just the method I'm used to applying when dealing with inequalities.
You need to go back and learn how to solve inequalities, and what it means to do so.

When you see 1<14x2<1-1 < \frac{1}{4}x^2 < 1 (which is correct), it doesn't mean you can pick one value in that interval. It doesn't even mean that you can find an x for which 14x2\frac{1}{4}x^2 will have any of those values.

It is just a statement that x can be any value for which that inequality is true. This is sort of the opposite of what you are saying.

I can't imagine how what you say can be a method you use to solve inequalities. Maybe you could show an example of how you do that.

14x2<1\frac{1}{4}x^2 < 1x2<4x^2 < 4x<2|x| < 22<x<2-2 < x < 2
The first line here happens to be equivalent to 14x2<1|\frac{1}{4}x^2| < 1, but only because the LHS can't be negative, so the absolute value is irrelevant: x2=x2|x^2|=x^2 by definition. You can't generally just drop an absolute value, but here you can. I think this is what you are saying.

The rest is okay, but you also have to be careful taking square roots of an inequality (the third line). You can't do that blindly, either. It is necessary to know that both sides are non-negative, so that the square root function is an increasing function.
 
Thanks. Got it. The next step is a bit more involved. I need to examine the behavior at the endpoints.

n=2(n!)2(2n)!22n+1\sum_{n=2}^{\infty}\frac{(n!)^2}{(2n)!}2^{2n+1}
n=2(n!)2(2n)!(2)2n+1\sum_{n=2}^{\infty}\frac{(n!)^2}{(2n)!}(-2)^{2n+1}
If I apply the ratio test, the result is 1, which is inconclusive. An alternative approach is to evaluate the limit directly.

limn(n!)2(2n)!22n+1\lim_{n\to \infty}\frac{(n!)^2}{(2n)!}2^{2n+1}
What method should I use to evaluate this limit?
 
First, check in each case whether the terms are all positive, alternating, or something else.

Then, consider the basics: divergence test, alternating series test, etc. If you were just given this series, noting the factorials, which tests would you try?
 
All the terms in the first series are positive, whereas all the terms in the second series are negative. The presence of factorials suggests applying the ratio test, but as I mentioned earlier, I already tried that and ended up with a limit of 1. Right now, I believe I’m using the divergence test, which involves evaluating the limit directly. I don’t have enough expertise to solve this limit fully, but the numerator provides some insight. It seems significantly larger than the denominator, which would imply the limit goes to infinity, though this is just a conjecture.
 
First, I assume you see that there is really only one question now, since the second is the negative of the first. So we're just looking at n=2(n!)2(2n)!22n+1\sum_{n=2}^{\infty}\frac{(n!)^2}{(2n)!}2^{2n+1}

I should state that I haven't tried until now actually working it out; I've just been encouraging you to show more work, and hopefully discover something! I have to say that I don't see the numerator as necessarily larger than the denominator, in part because I know how infinite things (here, products of very many factors!) can fool you. I can imagine factoring 2 out of half the factors in the denominator, and ultimately turning this into a quotient of products of very similar factors; I don't expect that to yield anything, so I'm not taking the time to write it out. But my sense is that the limit of the terms is likely to be 1 (not infinite), which of course would still make it divergent. (I also expect that since this is an endpoint, it ought to just barely miss converging, if at all.) But can I prove it?

One possibility would be to use Stirling's formula, which approximates the factorial. In fact, I just did that, and got a very interesting result, which agrees with what I found when I graphed the terms in Excel, to see whether they support my guess; but I'm either assuming you don't know about that, and can't use it, or would want to discover this for yourself. I'll just say that my guess above is not quite right.

Can you tell me where this particular problem came from? That might at least suggest whether some more advanced method like that is to be expected.

Or someone else may have a better idea.
 
This question is taken from the 4th edition of the Calculus textbook by Robert T. Smith and Roland B. Minton. It's Chapter 9, Exercise 16 from Section 9.6 on Power Series. The only difference is that the book uses the variable k, while I used n. I'll go back and review the chapter to see if there's any mention of Stirling's formula. Just to clarify, I'm not currently a student, and this isn't a particularly important question, so please don't feel the need to spend too much time on it.
 
WA notes
(n!)2(2n)!22n+1=2π  Γ(n+1)Γ(n+12) \dfrac{(n!)^2}{(2n)!}2^{2n+1} = 2\sqrt{\pi} \;\dfrac{\Gamma(n+1)}{\Gamma(n+\frac{1}{2})} which at least solves the divergence question. But if you look for approximations, you'll end up with Stirling. Stirling's formula has the big advantage that it provides an upper and a lower bound within strong limits.
 
Thanks. I'm not really sure what the functions on the right side are, I’m guessing they’re complex functions. After looking at the LaTeX code, it seems they’re Gamma functions. I’m not very familiar with how the Gamma function behaves, but I think Γ(n+1)>Γ(n+12)\Gamma(n + 1) > \Gamma(n+\frac{1}{2}). Does that mean 2πΓ(n+1)2\sqrt{\pi} \Gamma(n + 1) is significantly larger than Γ(n+12)\Gamma(n + \frac{1}{2}) and that the limit goes to infinity? Or am I misunderstanding something? I still haven’t found any mention of Stirling’s formula in the chapter. Is it inappropriate to use it if the chapter doesn’t cover it?
 
Thanks. I'm not really sure what the functions on the right side are, I’m guessing they’re complex functions. After looking at the LaTeX code, it seems they’re Gamma functions. I’m not very familiar with how the Gamma function behaves, but I think Γ(n+1)>Γ(n+12)\Gamma(n + 1) > \Gamma(n+\frac{1}{2}). Does that mean 2πΓ(n+1)2\sqrt{\pi} \Gamma(n + 1) is significantly larger than Γ(n+12)\Gamma(n + \frac{1}{2}) and that the limit goes to infinity?
Yes, to all. Except that the Gamma function can also be viewed as a real function. We have Γ(n+1)=n! \Gamma(n+1)=n! which can be generalized to a function of all real (and complex) values. The fraction in Γ(n+(1/2)) \Gamma(n+(1/2)) comes from the power of two term.
Or am I misunderstanding something?
No.
I still haven’t found any mention of Stirling’s formula in the chapter. Is it inappropriate to use it if the chapter doesn’t cover it?
It is the standard if you want to see how factorials behave because of its strong bounds. It is definitely worth looking up a proof of it. A quick search brought me https://kconrad.math.uconn.edu/blurbs/analysis/stirling.pdf But you need a bit of calculus, so I don't know whether it is appropriate for you or not.

If you only want to show that the terms increase monotonically, then a direct proof (or an induction) should work. Simply divide the n+1 n+1 -st term by the n n -th term and see whether it is greater than one or not.
 
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Thanks. I'm not really sure what the functions on the right side are, I’m guessing they’re complex functions. After looking at the LaTeX code, it seems they’re Gamma functions. I’m not very familiar with how the Gamma function behaves, but I think Γ(n+1)>Γ(n+12)\Gamma(n + 1) > \Gamma(n+\frac{1}{2}). Does that mean 2πΓ(n+1)2\sqrt{\pi} \Gamma(n + 1) is significantly larger than Γ(n+12)\Gamma(n + \frac{1}{2}) and that the limit goes to infinity? Or am I misunderstanding something? I still haven’t found any mention of Stirling’s formula in the chapter. Is it inappropriate to use it if the chapter doesn’t cover it?
You don't need to use the gamma function; you can use the Stirling approximation as stated in the link I supplied, and get a much nicer result.

I presume your textbook teaches all the basic convergence tests, so we should probably be looking either for one of them that can be applied more easily than finding this limit, or something about similar limits elsewhere in your book.

I just happens that the Stirling formula is mentioned in the textbook page I happen to have left open to check for any convergence tests I might be forgetting, but just in an exercise at the end of the previous section, when they state the formula and suggest using it to test convergence of a series closely related to yours. This suggests that this may well be the only method that will work. I hope you've checked the index of your book for Stirling's formula!

I searched for the book and found it in the Internet Archive. It doesn't mention Stirling, but I see that the problem is among many other "interval of convergence" problems, as if it is perfectly ordinary. So I assume there is an elementary way to solve it. We just have to keep thinking.

If you only want to show that the terms increase monotonically, then a direct proof (or an induction) should work.
This is a good idea.
 
I don't like nasi112\displaystyle 112 because she insulted me a long time ago when I was mario99\displaystyle 99. But there is no harm to show the audience:

limn(n!)2(2n)!22n+10\displaystyle \lim_{n\rightarrow \infty}\frac{(n!)^2}{(2n)!}2^{2n+1} \neq 0

We start with the denominator.

(2n)!=(2n)(2n1)(2n2)(3)(2)(1)\displaystyle (2n)! = (2n)(2n - 1)(2n - 2) \cdots (3)(2)(1)

The factors are either even\displaystyle \text{even} or odd\displaystyle \text{odd}, then:

(2n)!=[(1)(3)(5)(2n1)][(2)(4)(6)(2n)]\displaystyle (2n)! = [(1)(3)(5) \cdots (2n - 1)] \cdot [(2)(4)(6) \cdots (2n)]

We can factor the even\displaystyle \text{even} further.

(2)(4)(6)(2n)=2n[(1)(2)(3)(n)]=2nn!\displaystyle (2)(4)(6) \cdots (2n) = 2^n[(1)(2)(3) \cdots (n)] = 2^n n!

This gives:

limn(n!)2(2n)!22n+1=limn(n!)2[(1)(3)(5)(2n1)]2nn!22n+1\displaystyle \lim_{n\rightarrow \infty}\frac{(n!)^2}{(2n)!}2^{2n+1} = \lim_{n\rightarrow \infty}\frac{(n!)^2}{[(1)(3)(5) \cdots (2n - 1)]2^n n!}2^{2n+1}


=limnn!(1)(3)(5)(2n1)2n+1\displaystyle = \lim_{n\rightarrow \infty}\frac{n!}{(1)(3)(5) \cdots (2n - 1)}2^{n+1}


=limn2nn!(1)(3)(5)(2n1)2\displaystyle = \lim_{n\rightarrow \infty}\frac{2^n n!}{(1)(3)(5) \cdots (2n - 1)}2


=limn(2)(4)(6)(2n)(1)(3)(5)(2n1)2>0\displaystyle = \lim_{n\rightarrow \infty}\frac{(2)(4)(6) \cdots (2n)}{(1)(3)(5) \cdots (2n - 1)}2 > 0


@nasi112

Don't think that I forgot what you said, you F U C K E N moron😡
 
I have logistic_guy on "ignore," but I decided to see what this post was in reference
to the limit. I reported post #14 and requested a permanent ban because of the
major cursing and attack at the bottom. I am understanding that no one can ban
anyone anymore, though.

I was enjoying reading the mathematics in the lines of post # 14 until logistic_guy
told off nasi112 in that vulgar way.
 
=limn(2)(4)(6)(2n)(1)(3)(5)(2n1)2>0\displaystyle = \lim_{n\rightarrow \infty}\frac{(2)(4)(6) \cdots (2n)}{(1)(3)(5) \cdots (2n - 1)}2 > 0

The fraction can be written as

(2/1)*(4/3)*(6/5)*(8/7)* ... =

(1 + 1)*(1 + 1/3)*(1 + 1/5)*(1 + 1/7)* ...

Ignore the (1 + 1) factor for the next discussion.

(1 + 1/3)*(1 + 1/5) = 1 + 1/3 + 1/5 + 1/15 > 1 + 1/3 + 1/5

(1 + 1/3)*(1 + 1/5)*(1 + 1/7) > (1 + 1/3 + 1/5)*(1 + 1/7) =

1 + 1/3 + 1/5 + 1/7 + 1/21 + 1/35 > 1 + 1/3 + 1/5 + 1/7

Continuing this pattern,

(1 + 1/3)*(1 + 1/5)*(1 + 1/7)* ... *[1 + 1/(2n - 1)] >

1 + 1/3 + 1/5 + 1/7 + ... + 1/(2n - 1).

As n approaches infinity, this last sum is a divergent series.
 
I don't hate @lookagain but I wish that he apologizes to professor Stefan, even if privately. There is a reason why he has a very high ranking in my list! And I am sure that he will not let me down.

1 + 1/3 + 1/5 + 1/7 + ... + 1/(2n - 1).

As n approaches infinity, this last sum is a divergent series.
If anyone of you are wondering why this series diverges, you can use the comparison test.

n=112n1>n=112n\displaystyle \sum_{n=1}^{\infty}\frac{1}{2n - 1} > \sum_{n=1}^{\infty}\frac{1}{2n}

Since the smaller series n=112n=12n=11n\displaystyle \sum_{n=1}^{\infty}\frac{1}{2n} = \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n} diverges, then the bigger series will diverge as well.

And if you are wondering why the series n=11n\displaystyle \sum_{n=1}^{\infty}\frac{1}{n} diverges🤣you can follow my thread in the following post:
 
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