Is Convergence Possible for ejjej ejjej?

[Simkate]

ejjej

 

[Mark44]

Does this series convege or diverge- conditionally or absolutely ? ( with justification)

∑( ∞ to n=10) (-1)^n (2n)! / n!(2n)^n

I have use the RATIO TEST and after showing all my work i reached

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.
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lim(n-->∞) (2n/2n+1)
lim(n-->∞) 2/ (2+1/n)^n < 1

and therefore the series is convergent and is absolutely convergent


i don't know if my last two steps were correct, can someone help and make sure for me please.

I haven't checked all your work, but the limit in the next to last step is 1. That is, lim 2n/(2n + 1) = 1. I don't understand where your last step came from.

Please show what you ended with in the Ratio test.

 

[Simkate]

= (2n/2n+1)^n
so it is conditionally convergent?

 

[Mark44]

Here is my Work

∑( ∞ to n=10) (-1)^n (2n)! / n!(2n)^n

Using the Ratio Test
lim (n-->∞) [(-1)^n+1 (2n+1)! / (n+1!)(2n+1)^n+1] * [(n!) (2n)^n / (-1)^n) (2n)!]

Mistake above. The first fraction will be (2(n + 1))! /[(n + 1)! (2(n + 1))^(n + 1).

lim(n--->∞) [(2n+1)(2n)^n] / (2n+1)^n ( 2n+1)

lim(n--∞) [(2n)^n] / [ (2n+1)^n]

= (2n/2n+1)^n

Now i don't know what to after this...

i was wondering if its (n/n+1)^n --> (converges to) e >1

so it is conditionally convergent?

 

[Simkate]

What signaficance does the n=10 have?

the limit is still n to infinity right?



Thank you for that correction:

Now i have got:

after cancelling out terms through the ratio test
i ended up with

-lim(n-->∞) [(2n+2) (2n)^n] / [(n+1) (2n+2)^n]


-lim(n-->∞) [2n+2/ n+1] * [ 2n/2n+2]^n

I don't know what it converges to it is confusing me please help

 

[Mark44]

What signaficance does the n=10 have?

None that I can see. The series just happens to start at n = 10.

the limit is still n to infinity right?

Yes.

Thank you for that correction:

Now i have got:

after cancelling out terms through the ratio test
i ended up with

-lim(n-->∞) [(2n+2) (2n)^n] / [(n+1) (2n+2)^n]

That's not what I get, which is (2n + 1) nⁿ/(n + 1)^{n + 1}, which further simplifies to (2n + 1)/(n + 1) * [n/(n + 1)]ⁿ.

-lim(n-->∞) [2n+2/ n+1] * [ 2n/2n+2]^n

I don't know what it converges to it is confusing me please help

 

[Simkate]

Thank You i did see my silly mistake their and i have corrected now it makes sense.


so at the end

the lim (n--∞) 2n+1/n+1 * [ n/n+1]^n

i can divide all the terms by n?

to make it [ (2+1/n)/ (1+1/n) ] * [(1)/(1+1/n)^n] = 2/e < 1

therefore the series is absoluetely convergent?

 

[Mark44]

Yes, the series converges and is absolutely convergent. Good work! You picked up on the 2nd limit, which is 1/e.

 

[Simkate]

:) thank u so much i really appreciate it...it was so helpful