# 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) =...

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`

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### 1 Answer | Add Yours

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)`