I'm not sure what to do here, root test is = 1. Reddit series convergence test, r/calculus
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- เผยแพร่เมื่อ 28 ก.ย. 2024
- Calculus 2 series convergence test. Can we use the root test here? Or are there any tests we can use? This question is from Reddit r/calculus. Subscribe to @bprpcalculusbasics for more calculus tutorials. Check out my "100 series" video for more series convergence examples: • 100 series convergence...
Original post on Reddit r/calculus / sertu9pbsd
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ALWAYS do Divergence Test before doing any other test, UNLESS stated otherwise it’s already done. You could also do some other tests like p-series and telescoping when applicable.
The reason we just skip to the test in class, it’s because we skip problems or to the part where it passes the Divergence test, to accelerate the process, learning, or class, because DT is supposed to be a given to do so or given if it’s already done.
Such a neat problem. Got the nth term test for divergence to work by evaluating the required limit with L'Hopital's rule instead, but that was very messy. I like this approach too.
Immediately thought to use the divergence test since it looks like the limit should approach some power of e which can’t be 0 and lo and behold it does
"what even IS this?" is legitimately THE cue for doing the nth term test. if you had thought ratio test (a legit fallback after seeing the root test fails) you find yourself in a very similar line
Just fyi the ratio test is very much not a fallback after seeing that the root test fails -- the root test is strictly stronger than the ratio test, so if it fails, the ratio test will as well!
@@TheElihs you're absolutely correct of course. if the root test is _mechanically intractable_ you move on to the ratio test, but that's not the situation here. i loathe the root test and avoid it whenever it isn't trivial to apply, so i was clumsy handling this one.
Given n -> inf means getting arbitrarily close to a group of large numbers cant you logically conclude that since n - 3 is strictly < n + 1 (for reals) that the interior before the power is strictly < 1 meaning any power must also be strictly < 1 and > 0 meaning an infinite sum of non zero non significantly decreasing values which must diverge?
Why must the power be smaller then one in that case?
@@anezkabos6491 He's saying the interior raised to said power
Something always enrages me...
It approaches 0, not is.
meaning it will converge to 0 or diverge to infinity as the exponent tends to infinity.
it is only ever equal to 1 in the limit if the base is exactly 1
If π is an irrational number and you can find any sequence of digits in it you can find sequences like 00000, or 00000000 wouldn't that mean that π eventually has a sequence of zeros that approaches infinity in length making it rational?
A number with that property is called a normal number. It is not known whether π is normal though it is widely suspected that it is.
Even if it's true you can find a sequence of zeros of any length in the decimal expansion of pi (which is most likely true, just not proven yet), then it's not the same as having infinitely many zeros at its end. You can think of it as chunks of zeros being bigger and bigger forever, but all of them having ultimately finite length. It's like numbers getting bigger and bigger if you keep adding a constant, they tend to infinity, but never reach infinity, because infinity is not a number. Also, it's worth remembering the trailing zeros are only a consequence of a number being rational, not the definition - we know for a fact pi cannot be expressed as a ratio of integers as it's been proven in a number of ways.
@@pangadajski7687 but if 0.999... is equal to 1 doesnt that mean that 0.0000...1 is equal to 0, making every decimal after the infinite sequence of 0s of pi also equal 0, making it rational?
Yeah ok but can he figure out the infinite sum.
It diverges to positive infinite.
He shows that the infinith term of the series doesn't shrink to zero but to e^-4, which shows it grows to infinity.
That's why my teacher always said "test necessary condition of convergence first, it can rid you of long and complex calculations"
Nice
Nice
Surely a factorial is basically a constant? So n! Is also a constant? Why is n! not removed to the left of summation in root, ratio tests? I mean solving for n! in these tests are easy enough but needless … surely?
Why can you just say infinity/infinity equals 1 at the start? Isn't it indeterminate and you need to use l'hospital's rule?
Infinity/infinity=1? I thought it was indeterminate?
Lhopital
Note that we are taking the limit as x is approaching infinity of a rational function (polynomial divided by a polynomial) were the degree of he numerator and denominator are the same so we can take the limit to be the ratio of the leading coefficients.
In a more general case, factor out the highest power of x in both the numerator and denominator, cancel, and take the limit (usually infinity or 0).
Devi dare di nuovo l'esame
It is indeterminate, he just shortcut past L’Hôpital by noting since the top and bottom have the same greatest power and no coefficients, the limit is 1.
the term with the highest power (which is just n here) dominates in a rational function as n→∞
This is very cool haha, nice way of showing that the series diverges
Me being a 7th grader wondering why the answer is not (3/7)^8
n is an integer, which is different from a variable. The information you need is in the original equation where the big E looking thing (sigma) means we're taking the sum of all values between n=6 and n->inf. n being an integer means it increases by 1 in each of these values. So n = 6 then 7 then 8 etc.
Other people here can give you a much better explanation than this, but I hope this helped.
it helped quite alot , thank you@@eclectic6938
We could also just see that lim n->infinity(n-3/n+1) = 1. And since 1 raised to anything is still 1 and it is an infinite sum, it is basically the sum from 0 to infinity of 1 which is just infinity.
can you please evaluate this:
1-1+1-1+1-1+1-1+1-1+......=??
i tried my self and i came into 3 possible solutions:
-if number 1 is even, the solution is 0
-if number 1 is odd, the solution is 1
-if i take that sum = S then:
1-1+1-1+1-1+.... =S
1-(1-1+1-1+1-1+...)=S
1-S=S => 1=2S => S=1/2
So please can you tell us what is the correct answer?
the sum of that infinite series is 1/2
By TFD, the series diverges
If the number of terms are even then the sum is 0 and if the number of terms are odd then the sum is 1.
This sum is divergent (it does not converge to a value), in some contexts however it can be equal to 1/2 as
1/(1-x) = 1+x+x^2+x^3+x^4... for -1
Well it depends on what is meant by "evaluate".
The usual rules that apply to infinite sums imply that it diverges... TFD as in this video clearly shows that. In almost all contexts that's the correct answer.
HOWEVER (insert BIG caveat here) there is a way to assign a value to a divergent series using various means though this isn't a "sum" in the usual sense. That particular case can be handled by Cesàro summation which is one of the simpler menthods. That yields the value 1/2.
There is an excellent Mathologer video which describes this and includes the infamous 1 + 2 + 3 + 4 + ... example.
Given The Fact, couldn't you express (1-4/t)^t+1 as (1-4/t)^t x (t+1/t), and then given -4 as A and (t+1/t) as B, conclude that it is e^(-4t-4/t)? Why not?
I also thought of that first, but then i realised that t IS the limit itself, which means if B = (t+1)/t as t goes to infinity, then B = 1
If t + 1 is to be an exponent, you need grouping symbols around it.
why shave..