Homework Help

# Given the series  (x_n)_{n\in\mathbb{N^*}} \subset \mathbb{R^*_+} ...

emanuel- | Student, College Freshman | eNoter

Posted October 26, 2013 at 7:44 PM via web

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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



mlehuzzah | Student , Graduate | (Level 1) Associate Educator

Posted October 31, 2013 at 9:51 PM (Answer #1)

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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)]

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)

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