#help-41
1 messages ยท Page 14 of 1
,, \sum_{n=1}^{\infty} \frac{[(2n)!]^2}{[2^nn!]^4} \approx \sum_{n=1}^{\infty} \frac{2\pi (2n) \left (\frac{2n}{e} \right )^{2n}}{2^{4n}(2\pi n)^2 \left (\frac{n}{e} \right )^{4n}} = \sum_{n=1}^\infty \frac{e^{2n}}{\pi \cdot 4^n \cdot n^{2n+1}}
Bro but I don't think I could use the approximation
See
Bro let's leave it for a biy
bacc
Why
We can discuss this later
They are so close
that the quotient of n! and that formula is 1
that's how close they are
You are right
But the difficulty is that believe me
Its not that type of the assignment where u need Stirling
,w limit ( (e^(2n))/(pi4^nn^(2n+1) ))^(1/n) as n -> inf
root test implies it converges
wdym
you think there is a way more simple way
But root test was not conclusive for our original series
we tried like 5 tests
See
Use limit comparison test
See
We can discuss that later
Let's discuss the rest
So that I can complete them
I leave a page for 15.
Ok?
,w limit [((2n)!)^2/(2^nn!)^4] / [ (e^(2n))/(pi4^nn^(2n+1)) ] as n -> inf
wtf
Shi
yea
Can't use that nature
yea
โ ๏ธ
See
Bro
Let's discuss this later
We have 5 more questions
16-17-18-19-20

This question is not meant for this assignment but is given
Let's do the rest
so weird I expected 0
ok bro
16
,, \sum_{n=2}^\infty x^{n}\log^p n
bacc
root test seems inconclusive
Root test failed
for all p
Oof
My guess is that |x| < 1
D'alembert?
looking at this
Wait let's see
Then indeed
,w lim n tends to infinity [log(n)/log(n+1)]^q
|x| < 1
Its 1
ye
Lol
when it comes to power series
you can use these test
i think root test is cleaner
you can use squeze theorem with n^p
nothing
it fails naturally
cause then it is not even a null sequence
log(n)^p -> inf as n->inf
done
unless wait
p < 0
then you get 1/log(n)
If x=1 series will be [log(n)]^p
,w sum 1/(log(n))^1000
Lim n tend un
yea i am dumb
No no
So final
let's move on
Root test for basic
we got 2 minutes
And Positive term sequence test when x=1
ye
can you post it again 17
thanks g
No probs
ok 17 is kinda related to 15
but it's reciprocal
well not quiet
it's
,, \sum_{n=1}^\infty \frac{(n!)^2}{\left ( \frac{(2(n+1))!}{2^{n+1}(n+1)!} \right )^2}
Oof
we need to do actually
bacc
it starts with 3
Yes
Result:
3
ye
Yes
But
See
I feel in this root test alembert or rabbe might work
Like we might see 1/4 typa shi
I recommend root test if we have something with power to n
factorials are very good for ratio test
Okok
ok that's easy
,, \sum_{n=0}^\infty \frac{x^{2n}}{(n+2)\sqrt{n+1}}
x^(2n)
bet
Yes
bacc
probably ratio
ye
But I feel
then factor the highest n
Which would converge
Limit form
,w limit (1/[(n+2)(n+1)^(1/2)])/(1/[(n+2+1)(n+1+1)^(1/2)]) as n to infinity
Yea
ez
Yea
ok 19 ez
bet
yea
dont
yea
ratio
goes to 1
you can see you will have
n^(3/2) .... / n^(3/2)
so x^2 < 1
fr
fr
they dont sort from easy to difficult
they mix it
,, \sum_{n=1}^\infty \frac{n}{(2n-1)(2n)}
bacc
Fair
Ya fair now
1/n
Ez
Limit form
๐ฝ๏ธ
comparison test this
ye
This might be a good one
bet
bet
ratio test
no cap n^3/2 kinda attracting me
fr you got weird attrations
Lol
No prob
Bro
You the ๐
If the ๐ says one must listen
Ratio test
Ok
I believe on you
Fair we did all
Now we just need to crack the 15.
And we done
And bro pls also repeat the nth term for 17.
Then we done
Imma complete it
somewhere above
.
Fair
Ok we got
Now only we need to crack 15.
And we good to go
Then imma fair it out fast
bet
,w limit (n! - sqrt(2pi*n)(n/e)^n+1) as n -> inf
we were at this
Right
But applying approximation
And changing the nth term
Makes the root test conclusive now
This is the problem
I feel this will be done by direct comparison only
Or limit form
Sup
Sup
You've gotten the active role
Who?
You've grown
Who?
Check dm
The goat helping me
Aight gl
gl
fr
,, \sum_{n=1}^{\infty} \frac{[(2n)!]^2}{[2^nn!]^4}
.
bacc
Oof
,, \sum_{n=1}^{\infty} \left ( \frac{(2n)!}{2^{2n} n! \cdot n!} \right )^2 = \sum_{n=1}^{\infty} \left ( \frac{1 \cdot 2 \cdot ... \cdot (2n-1) \cdot (2n)}{2^{2n} n! \cdot 1 \cdot 2 \cdot ... \cdot (n-1) \cdot n} \right )^2
bacc
Yea
,, \sum_{n=1}^{\infty} \left ( \frac{(2n)!}{2^{2n} n! \cdot n!} \right )^2 \leq \sum_{n=1}^{\infty} \left ( \frac{(2n)^{2n}}{2^{2n} \cdot (n!)^2} \right )^2
bacc
,w limit ([(2n)^(2n)]/(2^(2n)*(n!)^2))^(2/n) n -> inf
So this diverges
yea
Positive term series
no
this just means
Oh shi I did not see 1/n
we found a too big upper bound
Got it
the comparison test would be valid if a greater series converges then ofc the smaller has too
Yes
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Perhaps try rephrasing your question?
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bro
Yes
one last test
Ok
,w limit ([(2n)^(2n)]/(2^(2n)*(n^n)^2))^(2/n) n -> inf
What is Vn
Auxiliary series
p series test?
,, \sum_{n=1}^{\infty} \left ( \frac{(2n)!}{2^{2n} n! \cdot n!} \right )^2 = \sum_{n=1}^{\infty} \left ( \frac{1 \cdot 2 \cdot ... \cdot (2n-1) \cdot (2n)}{2^{2n} \cdot 1^2 \cdot 2^2 \cdot ... \cdot (n-1)^2 \cdot n^2} \right )^2
And we compare this with that
bacc
What
gamma function
,w (2^(2 n) ฮ(1/2 + n))/(sqrt(ฯ) ฮ(1 + n))=(2n)!/(n!)^2
Yess
,, \sum_{n=1}^{\infty} \left ( \frac{(2n)!}{2^{2n} n! \cdot n!} \right )^2 = \frac{\sqrt{\pi}}{\pi} \sum_{n=1}^{\infty} \left ( \frac{\Gamma (\frac{1}{2}+n )}{\Gamma (1+n )} \right )^2
bet
nah i have it
See multiply and divide by gamma(1/2) in num and denom
It becomes beta(1/2,1/2+k)/gamma(1/2)
wait
wtf
ratio test works but it's not a null sequence
,w limit of Gamma(2n)/Gamma(1 + n) as n -> inf
bruh
Aah
bacc
Bro the term inside the bracket is literally
Beta(1/2,1/2+k)/gamma(1/2)
Gamma(1/2) take it out now you just have beta
,, \frac{\sqrt{\pi}}{\pi} \sum_{n=1}^{\infty} \left ( \frac{\Gamma (\frac{1}{2}+n )}{\Gamma (1+n )} \right )^2 =
\frac{\sqrt{\pi}}{\pi} \sum_{n=1}^{\infty} \left ( \frac{\Gamma (\frac{1}{2} + n )}{n\Gamma (n)} \right )^2
bacc
Bro
wait hold on
Sure
how
Beta(1/2,1/2+k)=gamma(1/2)gamma(1/2+k) whole by gamma(1/2+1/2+k) which is gamma(1+k)
Now see you have this only
In the bracket just multiply divide by gamma(1/2)
ok
,, \frac{\sqrt{\pi}}{\pi} \sum_{n=1}^{\infty} \left ( \frac{\Gamma(\frac{1}{2})\Gamma (n+\frac{1}{2})}{\Gamma(\frac{1}{2})\Gamma (n+1)} \right )^2
bacc
Yes
sorry lord oppailol
bacc
,, \frac{\sqrt{\pi}}{\pi^2} \sum_{n=1}^{\infty} \beta \left (\frac{1}{2},n \right )^2
bacc
Now?
but it's still a null sequence
๐โ ๏ธ
Lol
and see if that exists
Bro
nah
we are close
ok
Think logically
๐
In a assignment of convergence
Why would we have double integral including beta
Aah ๐โ ๏ธ
๐โ ๏ธ
It was the only way ๐โ ๏ธ
The problem is because of this I can't do other questions
This hindering me
,w integral Beta(1/2,x)^2 from x=1 to x=inf
so it diverges
Does not converge
by integral test
๐
I am cooked
you cooked
How I am supposed to summarise this whole chat on paper
deriving divergence with beta function
๐โ ๏ธ
Fr
Always 2 steps ahead
lmao
Help
you integrate a squared integral
I will be killed by a professor
!help
To ask for mathematics help on this server, please open your own help channel or help thread. See #โhow-to-get-help for instructions.
we already did
Its enough
No search the question
They will do some wrong shit
And get the answer
I will copy it lmao
Bro but yourself thino
Think*
I have an idea
,, \frac{\sqrt{\pi}}{\pi^2} \sum_{n=1}^{\infty} \beta \left (\frac{1}{2},n \right )^2 > \frac{\sqrt{\pi}}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{n}
ah no wrong
wait
bacc
try to prove this
๐
then done
This is tough than the question itself
Genuine question
shoot
,w Beta(1/2,x) > 1/x
โ ๏ธ
leave 1 page of tear drops
Assignment sheet bro
Lmfao
I am cooked tomorrow is the final deadline it will not extend now ๐โ ๏ธ
And I was not able to do this question
F it
I leave one whole page
Then I do 16-17-18-19-20
,, \beta \left (\frac{1}{2},n \right )^2 = \left (\int_0^1 \frac{1}{t^{1/2}}\cdot (1-t)^{n-1} \dd t \right )^2
bacc
Lol
this could work
๐
Imagine
Such type of a question in an assignment which is considered easy
The question is wrong
you are wrong
๐
I am leaving a page for it
skewed ass sheet
Tomorrow will be a debate
is it that bad if you wont complete just 1
Between me and him
Na 17 is chill
pls tell me
did we?
Yes
i thought we skipped
dont remember
i said it lol?
diverges?
ah yea think limit was 4
Yes
you are my hero
๐
See every question done till now has taken Max to max 1.5 pages
That's why I left 1 page accordingly
I can't do in 3 pages
A single question
@patent raptor
Give the nth term
Pls
huh
.
2(n+1)
yea
In power 2 ?
we said it starts at 3 so n+1 instead n
In power of 2 it is also n+1?
,calc (1/2)^2
Result:
0.25
,calc (13)^2/(24)^2
Result:
0.140625
,, \left ( \frac{1}{2} \right )^2 + \left ( \frac{3}{8} \right )^2 + \left ( \frac{5}{16} \right )^2 +\left ( \frac{35}{128} \right )^2 + ...
bacc
I had hopes to detect a pattern
nothing i analyzed 15
Oof
can you see a pattern
In the denominator
Anything is fine
$\textbf{15.}\$
[ \sum_{n=1}^\infty \prod_{k=1}^n \left ( \frac{2k-1}{2k} \right )^2 ]
bacc
Now the question is if we can do anything with products
,w prod (2k-1)^2/(2k)^2 from k = 1 to n
Ik
The most probable answer is
The question is wrong
It was not supposed to be this
but at least
at least we found out it's diveregnt
because of the beta function
graphically speaking
i would honestly do it like this with the beta function and argue that graphically 1/n is less and by comparison test it is divergent
supposed to be 4
Okok
i assumed you did a_n+1 / a_n
fr
15 is impossible
calc 2 rn?
Without higher implementation
Yes
Marks will be cut from internal
If I submit incomplete
except if there is some trick
i searched internet but no success
.
Aah right
,w double factorial
๐โ ๏ธ
Wait let's see
I mma write this derivative and results
In a rough page
Then will show him
Tomorrow
I am sure question is wrong
,, \sum_{n=1}^{\infty} \left ( \frac{(2n)!}{(2^n n!)^2} \right )^2 = \sum_{n=1}^{\infty} \left ( \frac{(2n-1)!!}{2^n n!} \right )^2 > \sum_{n=1}^{\infty} \left ( \frac{(2n-1)!}{2^n n!} \right )^2
relax
Ok
,w limit (2n-1)!/(2^nn!) n -> inf
do this
yes
I have left one page for this
Lol
look it up
Which should be evident itself from when we were doing
yea
Then why was it showing 0
,w limit (2n)!/(2^nn!)^2 n -> inf
,w plot y = (2x)!/(2^x x!)^2 and y = (2x-1)!!/(2^x x!) between x = 0 and x = 10
This is guiding us wrong
bro
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Ok
it's wrong anyway I think then
Same
Cz it diverges
If you can express simply
I can do this in 2 lines
Cz now we know it diverges
So lim n tend to inf nth term should not be zero
,w Sum[((2n)!/(2^(2n)(n!)^2))^2,{n,1,100}]
inf doesnt work
but yea it diverges 100 %
finally
Either we derive something with this
bacc
We can only tell that n! is in (2n)!
So if we were to cancel things we would have some terms of (2n)! left
bacc
\begin{gather*} = \sum_{n=1}^{\infty} \left ( \frac{(n+1) \cdot (n+2) \cdot ... \cdot (n+n-1) \cdot (n+n)}{2^{2n} n!} \right )^2
\end{gather*}
bacc
I wonder if we wouldn't have technically n^n in the numerator
bro
Other alternative?
post 20
,w limit [ (log(n+1)^2)/((n+1)^(3/2)) ]/[ (log(n)^2)/((n)^(3/2)) ] n ->inf
Its still failing
What should we do
@patent raptor
Idea came
Bacc
Bacc
What if we do integral test
?
Aah
Bro played with me
Fr
But my guy 1/n^3/n^3/2 will converge
With which
Aah
.
shit
should still work with 1/3