Given the series 

`(x_n)_{n\in\mathbb{N^*}} \subset \mathbb{R^*_+} ` and `(y_n)_{n\in\mathbb{N^*}} \subset \mathbb{R^*_+}` with

 `(x_n) = \sum_{k=n}^{2n-1} ln^3(1+\frac{1}{k})`` `

and

`(y_n) = [ln(n+1)-ln(2n+1)+ln2]\; ln\frac{n+1}{2}ln(1+\frac{1}{2n^\alpha})\;, \forall n \in \mathbb {N^*}`

a)Show that Xn is convergent, and that its limit is 0.

b)Determine, in terms of `\alpha` , what kind of series `y_n` is.

c)For `\alpha` = 1, show that `y_n = \frac{1}{3}(x_n-x_{n+1}), \forall n \in \mathbb{N^*}` , which you will then use to find the sum `\sum_{n=1}^{\infty}y_n`

` `

Expert Answers

An illustration of the letter 'A' in a speech bubbles

First, observe that `x_n` is positive.  So, if we can show that `x_n` is smaller than a sequence that goes to 0, then `x_n` must also go to 0.

As k increases, 1+1/k decreases, so ln(1+1/k) decreases

Thus, if `n<=k` , then `ln(1+1/n) >= ln(1+1/k)`  

So,

`ln^2 (1+1/n) ln(1+1/k) >= ln^3(1+1/k) `

and we can write

`x_n <= ln^2(1+1/n) sum_(k=n)^(2n-1) ln(1+1/k)`

we can take this sum using facts about logs:

`ln(1+1/k)+ln(1+1/(k+1))+...=ln( (1+1/k)(1+1/(k+1)) *...)`  

But that product will always be 2: most of the numerators and denominators will cancel:

`(1+1/k)(1+1/(k+1))...(1+1/(2k-1))`

`=((k+1)/k)((k+2)/(k+1))((k+3)/(k+2))...((2k)/(2k-1))`

`=((2k)/k)=2`

So,

`x_n <= ln^2(1+1/n) ln(2) `

`<ln^2(1+1/n)`

Now, as `n->oo` , `1+1/n ->oo` , and `ln^2(1+1/n)->0`

So `x_n->0`

I can't actually understand what is written for `y_n` , it didn't render properly.  But here are some things to consider for part c:

`x_n-x_(n+1)=`

`sum_n^(2n-1) (...) -sum_(n+1)^(2n+1) (...) =`

`ln^3(1+1/n)-ln^3 (1+1/(2n)) - ln^3 (1+1/(2n+1))`

Think of this as `x^3-y^3-z^3` , where

`x=ln((n+1)/n)` , `y=ln((2n+1)/(2n))` , `z=ln((2n+2)/(2n+1))`

Then `y+z=ln[((2n+1)/(2n))( (2n+2)/(2n+1))]=ln[(2n+2)/(2n)]=ln((n+1)/n)=x`

So now the goal is to rewrite everything we can as (y+z), and then turn it into x:

`x^3-(y^3+z^3)=x^3-(y+z)(y^2-yz+z^2)=`

`x^3-x(y^2-yz+z^2)=x^3-x(y^2+2yz+z^2-3yz)=`

`x^3-x((y+z)^2-3yz)=x^3-x(x^2-3yz)=`

`x^3-x^3+3xyz=3xyz`

So: `1/3 (x_(n+1)-x_n)=xyz=`

`ln((n+1)/n)ln((2n+1)/(2n))ln((2n+2)/(2n+1))=`

`ln((n+1)/n)ln(1+1/(2n))[ ln 2 + ln(n+1)-ln(2n+1)]`

IF you had

`y_n=1/3 (x_n-x_(n+1))`

(which I am not sure about, since, as I said, `y_n` does not appear to have displayed properly)

THEN:

`sum_1^oo y_n = 1/3 x_1`

since all the other terms are subtracted off:

`(x_1-x_2)+(x_2-x_3)+(x_3-x_4)+...=x_1`

(provided the terms go to 0)

`1/3 ln^2(1+1/1)=((ln 2)^3)/3`

You can rewrite `x^3-y^3-z^3` as `x^3-(y+z)(y^2-yz+z^2)`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial