what to do if the second derivative test is inconclusive
Ratio test inconclusive, what now?
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Homework Statement
Determine departure/convergence of this series:
[itex]\sum[/itex][itex]\frac{(2n)!}{(due north!)^{2}}[/itex] * ([itex]\frac{1}{4}[/itex])[itex]^{n}[/itex]
Homework Equations
ratio test?
The Attempt at a Solution
I attempted to use the ratio exam and the resulting limit was 1, which means the ratio test is inconclusive. So far, I have learned to utilise only the ratio examination with series involving factorials and powers (of n); even so I have also been taught the other common dvgs/svgs tests (root, alternating, integral, etc).
I honestly take no thought on where to get from hither. I tried researching online, only I only establish references to sup and inf limits and I don't even know what those are. 
Answers and Replies
The ratio examination is conclusive with me. Could yous post your calculations??
It looks inconclusive to me. I get the ratio to be (2n+2)*(2n+ane)/((north+1)*(north+1)*four). Then ratio goes to one. Or am I doing information technology wrong too?
It looks inconclusive to me. I get the ratio to be (2n+2)*(2n+1)/((n+1)*(northward+1)*4). So ratio goes to 1. Or am I doing information technology wrong too?
Aaaargh, no, I made a mistake. Information technology is inconclusive
Thank you Dick!Homework Statement
Determine divergence/convergence of this series:
[itex]\sum[/itex][itex]\frac{(2n)!}{(n!)^{2}}[/itex] * ([itex]\frac{1}{4}[/itex])[itex]^{n}[/itex]
Homework Equations
ratio exam?
The Attempt at a Solution
I attempted to use the ratio examination and the resulting limit was one, which means the ratio exam is inconclusive. And so far, I have learned to use merely the ratio test with series involving factorials and powers (of due north); still I take too been taught the other common dvgs/svgs tests (root, alternate, integral, etc).I honestly have no idea on where to get from here. I tried researching online, but I only establish references to sup and inf limits and I don't even know what those are.
I would pull out a version of Stirling's approximation and endeavour to figure out what the series really looks like for big due north.
I would pull out a version of Stirling's approximation and try to figure out what the series actually looks similar for large n.
What exactly is Stirling's approximation?
[tex]L = \lim_{yard \rightarrow \infty} k\left(\frac{a_k}{a_{m+1}} - one\correct)[/tex]
If the limit exists, then L < 1 implies convergence, and L > 1 implies deviation of the series
[tex]\sum_{k=i}^{\infty}a_k[/tex]
L = 1 is inconclusive.
Raabe'due south test is a special case of Kummer's test. The proof is pretty easy. Encounter, eastward.g., Thomson, Bruckner, and Bruckner, Unproblematic Real Assay.
Then I would substitute the stirling approximation for every due north!, correct?
Try information technology! jbunniii'southward suggestion of the Raabe examination is a nifty idea too.
, applied Raabe's exam and got -[itex]\frac{1}{2}[/itex], therefore the series is divergent, right?I too put the sum into WolframAlpha using the Stirling approximation and it just states that the series does not converge.
[tex]\log a_n = \log\left(\frac{(2n)!}{(n!)^two} \left(\frac{1}{iv}\correct)^n\right)[/tex]
This allows yous to simplify the expression significantly. You lot know that convergence is impossible unless
[tex]\lim_{n \rightarrow \infty} a_n = 0[/tex]
or equivalently
[tex]\lim_{n \rightarrow \infty} \log(a_n) = -\infty[/tex]
If this isn't truthful, then you know immediately that the series diverges.
Alright, and then I, hopefully correctlyI also put the sum into WolframAlpha using the Stirling approximation and it merely states that the series does not converge.
I got +one/2, not -1/ii, only I might have fabricated a mistake. If you post your work, I'll compare notes with you.
P.Southward. You probably already noticed, but I got the inequalities reversed in my postal service almost Raabe's exam. L > 1 implies convergence, and Fifty < 1 implies divergence. And then either +1/2 or -1/2 requite the same conclusion.
I found a more elementary solution, assuming I didn't brand a fault. Consider taking the log of each term:[tex]\log a_n = \log\left(\frac{(2n)!}{(due north!)^2} \left(\frac{one}{iv}\right)^n\right)[/tex]
This allows you to simplify the expression significantly. You know that convergence is impossible unless
[tex]\lim_{northward \rightarrow \infty} a_n = 0[/tex]
or equivalently
[tex]\lim_{due north \rightarrow \infty} \log(a_n) = -\infty[/tex]
If this isn't true, and so yous know immediately that the series diverges.
If you apply the Stirling approximation you tin can conclude the limiting behavior of the series proportional to a ability of n for large n. Or yous tin estimate this by using numerical experiments. They'll show you it does approach 0. Simply like what power? Answering that question will tell y'all whether the serial converges.
I institute a more elementary solution, assuming I didn't make a mistake.
Sure enough, I made a mistake - scratch that. The terms go to nil, as Dick said.
Later a bit more scribbling, I found an elementary argument that works. It turns out to exist quite elementary, actually.
Write out the factors of [itex](2n)![/itex]
The product of the even factors is [itex]2^n n![/itex]. The product of the odd factors is
[tex](2n-one)(2n-iii)\cdots three \geq (2n-2)(2n-4)\cdots2 = ii^{n-i}(n-i)(n-2)\cdots one = 2^{north-ane}(due north-1)![/tex]
Thus
[tex](2n)! \geq ii^n 2^{north-1} n! (north-i)![/tex]
from which you tin can easily bear witness that
[tex]\frac{(2n)!}{(due north!)^ii} \frac{ane}{4^n} \geq \frac{1}{2n}[/tex]
and thus the series diverges by comparison with the harmonic series.
Sure enough, I fabricated a fault - scratch that. The terms go to zip, as Dick said.After a bit more scribbling, I found an simple argument that works. Information technology turns out to be quite elementary, really.
Write out the factors of [itex](2n)![/itex]
The product of the even factors is [itex]2^n due north![/itex]. The product of the odd factors is
[tex](2n-ane)(2n-iii)\cdots 3 \geq (2n-two)(2n-4)\cdots2 = 2^{due north-1}(northward-one)(northward-ii)\cdots ane = two^{northward-1}(n-1)![/tex]
Thus
[tex](2n)! \geq 2^n 2^{north-1} n! (n-1)![/tex]
from which you can easily evidence that
[tex]\frac{(2n)!}{(n!)^2} \frac{1}{four^n} \geq \frac{i}{2n}[/tex]
and thus the series diverges by comparison with the harmonic series.
Very overnice. Stirling will give you the more accurate estimate that series is asymptotically c/sqrt(n). So it diverges by a limit comparing test. But I like your elementary approach.
Very nice. Stirling will give you lot the more authentic estimate that series is asymptotically c/sqrt(n). So it diverges by a limit comparison test. But I like your unproblematic approach.
Yes, Stirling's approximation is handy here, but I was a bit puzzled that the problem would call for information technology or for Raabe's test when evidently these hadn't been covered nevertheless in the OP'due south studies, and then I was happy to find a simple solution. Plus it's a nice play a joke on that could come in handy once more onetime, thus becoming a method instead of a trick.
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