Determine if the alternating series converges or diverges.
>>7690216
Can't do it, not even if sober.
The series diverges
>>7690219
ur kidding right? it clearly converges
Why are you in a lab coat, weirdo? You're doing math. Wear Khakis and polo shirts like the rest of us.
>>7690229
It clealy doesn't. Are you retarded?
>>7690233
http://www.calcchat.com/book/Calculus-10e/9/5/23/
>>7690231
Because labcoats look dedicated and professional, and with the right features, quite sexy.
The image's template is based around vaguely taunting and expectant sexiness.
>>7690216
it converges
to
1/3 + 2/(3 srt(3)) arccsch(sqrt(2))
>>7690362
https://www.wolframalpha.com/input/?i=sum+%28%28-1%29^%28n%2B1%29*%28n!%29%29%2F%282n-1%29!!%2C+n%3D1+to+infinity
>>7690216
Did no one noticed the fault in notation?
Sci is getting more fucked every day.
>>7690374
>Girl look at dat denominator
>>7690216
To solve write as sum of (-1)^(n+1) 2^n/(2n,n).
You can write 1/(2n,n) as an integral then move sum inside integral/geometric series yadda yadda to get: int t from 0 to 1 of 2t/(1+2t-2t^2)^2 dt.
Solve that using partial fraction decomposition and so on to get (1+log(2+√3)/√3)/3
>>7690381
Finally some one else who noticed and posted.
But really though I can't fucking do this number theory bullshit, I'm not some Cleo on stackexchange
http://www.wolframalpha.com/input/?i=sum+of+%28-1%29%5E%28n%2B1%29n%21%2F%282n-1%29%21%21+from+n%3D1+to+infinity
Also, as n gets big the ratio becomes 1/2, so like duh
>>7690391
This is term 1 analysis, not number theory.
Here's a better question:
Schrodinger has one deutschmark. He is standing in front of 100 boxes, each of which
contains either a dead cat or an alive cat. When he comes up to a box, he bets some money and
makes a guess about the state of the cat (he can make any bet, even a non-integer one, but it
can’t be larger then the money he has got). If he guesses correctly, he doubles his bet, if not –
loses it. What is the maximal possible capital he can guarantee for himself after the 100 boxes,
given that he knows:
a) There is exactly one dead cat;
b) There are exactly n dead cats (where n ∈ [0, 100]).
>>7690374
Why the fuck is there a !! in your denominator?
>>7690419
double factorial means multiplying down like a factorial but you skip a number each step. That gets us 1*3*5*...(2n-1)
>>7690216
Diverge
>>7690380
What fault?
>>7690422
Is there something I'm missing here with the notation? I don't see any reason for there to be a !! in the original problem.
>>7690216
We have n+1/n * (2n-1)/(2n+1) < 1
Thus the absolute values are decreasing so the series converges.
>>7690374
There is no double factorial...
(2n-1) is the denominator
n=1, denominator = 1
n=2, denominator = 3
n=3, denominator = 5
...etc
It's just a stupid spiel on series notation.
Ratio test kills this problem in an instant.
It's convergent.
>>7690431
is shows the first results in the denominator 1,2,3 and then does the (2n-1).
In sum notation you only have to use the (2n-1)
Is twice the same thing.
>>7690443
Isn't this what OP intended though?
>>7690216
All Alternating series that have the absolute value of the successive terms approaching 0 converges, so, this clearly converges. This is true because the sequence of even n partial sums is a monotonically increasing sequence, and the sequence of odd n partial sums is a monotonically decreasing sequence, where every odd partial sum is an upper bound of every even partial sum, and every even partial sum is a lower bound of every odd partial sum (if the odd partial sum has more terms than the even partial sum, then their difference is e-o+e-o...+e-o+e, (+), if the odd partial sum has fewer terms than the even partial sum, then the difference is -(-o+e-o+e...-o+e-o), (+)). Monotonically increasing sequences of partial sums bounded from above converge to a real number, and similarly for monotonically decreasing sequences bounded from below, and because the absolute value of each term goes to 0, the limits are the same.
>>7690460
As a joke maybe.
Either way, notation is wrong.
>>7690443
You're misreading it, it indicates a chain product of numbers 1*3*5*...*(2n-1), nothing redundant. The notation OP uses is wrong of course, so you could probably interpret it that way too, idk.
>>7690464
I don't think it was as a joke.
If it is as you suggested, then the question is too easy and not worth making a thread over.
>>7690462
*where the absolute value of successive terms approaches 0 monotonically
>>7690466
So if it's a double factorial on the bottom then it converges, but if it's not and it's just shitty notation then it diverges? Fun stuff
>>7690490
Ambiguous notation will guarantee that this thread won't die.
>>7690362
>2n - 1
Wrong sequence kiddo.
First two terms are the same, but the 3rd term onwards is wrong.
Also double factorial is indeed a correct way to express 1 x 3 x ... x (2n - 1)
>>7691828
Forgot pic. Compare it to OP pic.
LOL. Answer converges. Bottom is just Gamma(n+1/2)*2^n/Sqrt(Pi). Converges. No time to figure out closed form, probably shit with inverse cos/sin, dont really give a shit about this one though
>>7690216
diverges
this is basic math shite
>>7690216
anyone who says it converges needs to repeat 10th grade.
This is not number theory.
This is not integral calculus.
Hell, this is not even series mathematics.
None of that is needed,
This is pure, fucking, power rules and logic.
-1 to the n'th power, times -1 time n factorial is the numerator.
the denominator is double factorial, but it reaches 2n-1 so it is ABSOLUTELY equal to n factorial.
when n approaches infinity, n factorial and the denominator cancel out, leaving only:
-1 to the n'th power times -1
so each term in this series will always be equal to either 1 or -1.
Each two subsequent terms cancel out
The sum of the series is 0
>>7691906
anyone who says it diverges*
>>7691906
>so each term in this series will always be equal to either 1 or -1.
>Each two subsequent terms cancel out
>The sum of the series is 0
If the terms would look like 1,-1,1,-1,1,-1,... the series would diverge.
>>7691930
yeah, my bad, q is -1 so it's a regular diverging geometric series so to speak
>>7691936
but actually i'm wrong all together
this shit converges
just put different Ns in the expression
looks like i need to repeat some classes
Real answer: you can't know whether or not it diverges or converges only by looking at it for seconds. You have to work out (something that nobody has done so far) in order to reach the solution or to use a software
>>7690384
Looking at this integral again you can solve it more easily. Complete the square
and substitute to get:
2√3/9 int t from -1/√3 to 1/√3 of (√3 t+1)/(1-t^2)^2 dt
Then subsitute t=tanh(t') and tidy to get:
1/9 int t from -atanh(1/√3) to atanh(1/√3) of (3 sinh(2t) + √3 cosh(2t) + √3) dt
From there it's basic integration/trig.
You end up with:
1/9 (3 + 2 √3 atanh(1/√3))
Now I will work it out for you.
Look at the denominator. That is the multiplication of all the odd numers less than a the double of a given number n. For example, n=10:
1*3*5*7*9*11*13*15*17*!9
The general formal is given by:
(2n)!/(n! * 2^n)
We now replace (2n)!/(n! * 2^n) in the denominator. Then we multiply n! * 2^n in the denominator and the numerator and we get pic related.
Now, let's think about it. n!*n!*2^n is clearly much bigger than (2n)! so the fraction approaches infinity as the value of n is bigger.
SInce the sign is determined by (-1)^(n+1), for odd values of n+1 we will have a subtraction and for even values of n+1 we will have a sum.
But since the fraction itself never converges, with (-1)^(n+1) will not either converge. The summation will alternate sums and substractions of huge number. The sign of the result will be negative for odd values of n+1 (even values of n), which would yield in the end a huge negative number. And for even values of n+1 (odd values of n) the summation would yield a huge positive number. So the summation diverges.
https://www.wolframalpha.com/input/?i=sum+%28%28-1%29^%28n%2B1%29*%28n!%29%29*%28n!+*+2^n%29%29%2F%282n%29!%2C+n%3D1+to+infinity
Kek so it converges in the end. The more you know.
