2 questions - series & inequalities

[KLscilevothma]

2 questions -- series & inequalities

1. By differentiate the function 1/(1-x), or otherwise, show that

inf
[sum] n²/2ⁿ = 6
n=1

2) Given Holder's Inequality http://mathworld.wolfram.com/HoeldersInequalities.html(equation 4)
show that (attached file)

 

[KLscilevothma]

exam questions

The attatchment is ready now. These 2 questions appeared in my exam but I don't know how to do them.

 

[metacristi]

1.I do not know whether my solution is of help for you (it implies the basic knowledge of linear inhomogeneous recurrence) anyway I will present it.

Let

S=∑_n=1^∞ n²/2ⁿ (1)

and

S1(n)=∑_k=1ⁿ k²/2^k (2)


with S=lim_n->∞ S1(n) (3)


We have (equalities 3'):

n=1 ---> S1=1/(2¹)

n=2 ---> S1=6/(2²)

n=3 ---> S1=21/(2³)

n=4 ---> S1=58/(2⁴)

n=5 ---> S1=141/(2⁵)

n=6 ---> S1=318/(2⁶)

Let now the general term of the sequence 1=a[1],6=a[2],21=a[3],58=a[ 4 ],141=a[5],318=a[6]... be a[n].

The closed form for the sum S1(n) is:

S1(n)=a[n]/(2ⁿ) (4)

To find a[n] it must be observed that (equations 4'):

n=1 ---> a[1]=1

n=2 ---> a[2]=6

n=3 ---> a[3]=21

n=4 ---> a[ 4 ]=58

n=5 ---> a[5]=141

n=6 ---> a[6]=318

and

a[2]-a[1]=1*2+3=5

a[3]-a[2]=5*2+5=15

a[ 4 ]-a[3]=15*2+7=37

a[5]-a[ 4 ]=37*2+9=83

a[6]-a[5]=83*2+11=177

a[7]-a[6]=177*2+13=367

a[8]-a[7]=367*2+15=749

........

a[n]-a[n-1]=2*{a[n-1]-a[n-2]}+(2n-1) (5)


If you have some experience with inhomogeneous sequences it's easy to find that we must seek for the closed form of a[n] a formula of the type:

a[n]=[A*n²+B*n+C]+D*2ⁿ

Introducing in (5) the equations 4' for n=1,2,3,4 and solving the system --->

A=-1
B=-4
C=-6
D=6

Therefore

a[n]={-1*n²-4*n-6}+6*2ⁿ (6)

Introducing (6) in (4) --->

S1(n)={{-1*n²-4*n-6}+6*2ⁿ}/(2ⁿ)} (4)

Finally

S1(n)=6-{(1*n²+4*n+6)/(2ⁿ)

and

S=6

 

[metacristi]

I don't understand what happened I used the edit feature...

 

[metacristi]

I don't understand what happened I used the edit feature...

 

[metacristi]

I don't understand what happened I used the edit feature...sorry.

 

[Hurkyl]

1/(1 - x) = Σ_n=0..∞ xⁿ

Differentiate both sides of the equation with respect to x, see if that gives you any clues. What if the numberator was just n instead of n squared; could you use the derivative you just computed to evaluate the sum?


I haven't worked through the second one before... I'll see if I can give a hint once I have time to figure it out.


Edit: changed the lower case sigma to upper case sigma

 

[KLscilevothma]

Thanks for reply.
metacristi, your method is pretty new to me but I'll try to understand that.

Originally posted by Hurkyl
1/(1 - x) = σ_n=0..∞ xⁿ
Differentiate both sides of the equation with respect to x, see if that gives you any clues.

The question clears up a bit now. I'll try to work it out after my calculus exam.

For question 2, please take your time, there's no rush. I'll try to work out the answer 2 days later, after my exams. Again thanks.

 

[KLscilevothma]

I'm terribly sorry

that I forgot to mention that a₁,a₂...a_n>0 such that [sum]_r=1..n a_r = 1 for the second question.

 

[metacristi]

2.From what I've seen Holder's inequation hold only for (1/p)+(1/q)=1 and p,q >1 so that,I suppose,these are assumed true in the case of your problem too.

Let

A_r=(a_r+1/a_r)

B_r=n^-1/q

We have A_r,B_r > 0 and B_r <= 1 (if q > 1).

∑_{r=1 to n} A_r*B_r=∑_{r=1 to n} n^-1/q* (a_r+1/a_r) (1)

[∑_{r=1 to n} (A_r)^p]^1/p=[∑_{r=1 to n} (a_r+1/a_r)^p]^1/p (2)

[∑_{r=1 to n} (n^-1/q)^q]^1/q=(n*1/n)^1/q=1 (3)

From Holder inequation,taking into account (1) (2) & (3) --->

[∑_{r=1 to n} (a_r+1/a_r)^p]^1/p >= ∑_{r=1 to n} A_r*B_r=∑_{r=1 to n} n^-1/q* (a_r+1/a_r)

We have a_r+1/a_r >= 2--->

∑_{r=1 to n} (a_r+1/a_r)^p >= [∑_{r=1 to n} n^-1/q* (a_r+1/a_r)]^p=n^-p/q*[∑_{r=1 to n} (a_r+1/a_r)]^p

 

[KLscilevothma]

Thanks.

Originally posted by metacristi
Let

A_r=(a_r+1/a_r)

B_r=n^-1/q

We have A_r,B_r > 0 and B_r <= 1 (if q > 1).

&#8721_{r=1 to n} A_r*B_r=&#8721_{r=1 to n} n^-1/q* (a_r+1/a_r) (1)

[&#8721_{r=1 to n} (A_r)^p]^1/p=[&#8721_{r=1 to n} (a_r+1/a_r)^p]^1/p (2)

[&#8721_{r=1 to n} (n^-1/q)^q]^1/q=(n*1/n)^1/q=1 (3)

From Holder inequation,taking into account (1) (2) & (3) --->

[&#8721_{r=1 to n} (a_r+1/a_r)^p]^1/p >= &#8721_{r=1 to n} A_r*B_r=&#8721_{r=1 to n} n^-1/q* (a_r+1/a_r)

We have a_r+1/a_r >= 2--->

&#8721_{r=1 to n} (a_r+1/a_r)^p >= [&#8721_{r=1 to n} n^-1/q* (a_r+1/a_r)]^p=n^-p/q*[&#8721_{r=1 to n} (a_r+1/a_r)]^p

I don't understand the second last step. Why do we need a_r+1/a_r >= 2. (I know it can be proved by AM>=GM)

In fact I found out how to do it yesterday but didn't have time to post my proof. I used a similar approach and here's my proof.
http://www.angelfire.com/freak2/anywork/ine2.jpg

From what I've seen Holder's inequation hold only for (1/p)+(1/q)=1 and p,q >1 so that,I suppose,these are assumed true in the case of your problem too.

Actually it is part (b) of a long question. I think I better post the whole question here.
http://www.angelfire.com/freak2/anywork/inequality.jpg
I am going to do part (d) today, and will ask again if I can't do it.

 

[bogdan]

I think I got it...but it's quite simple...

if you know a)...then just put from 1/p+1/q=1, q=p/(p-1)...in d)...

sum[(Ar+1/Ar)^p]>n^(1-p)*[1+sum(1/Ar)]^p;
prove that sum(1/Ar)>n^2 this way...

sum(1/Ar)>n*(fact(1/Ar))^(1/n)=n*1/[fact(Ar)]^(1/n)>

> n/[sum(Ar)/n]=(n^2)/1...that's all...I think...

 

[bogdan]

I think I have a solution for a)...
a) is equivalent to

{{sum[(Ar+1/Ar)^p]}^(1/p)}*n^(1/q)>sum(Ar+1/Ar);

write n this way ...

n=sum(1^q)...

{{sum[(Ar+1/Ar)^p]}^(1/p)}*{[sum(1^q)]^(1/q)}>sum(Ar+1/Ar);

which is exactly holder ineq for (Ar+1/Ar) and (1)...(hope it's correct...

 

[metacristi]

I don't understand the second last step. Why do we need ar+1/ar >= 2. (I know it can be proved by AM>=GM)

I intended to underline the fact that a_r+1/a_r > 0 but we don't need it,indeed,in the demonstration since a_r are all positive.

 

[metacristi]

To prove a. suffices to write q as:

q=p/(p-1) and by studying the function q=f(p) --->

-when p < 0 ---> q is always greater than 0 ---> pq < 0.

-when p belongs to (0,1) ---> q is always negative ---> pq < 0.

-when p belongs to (1,+∞) that's it when p > 1 we have q greater than 0 ---> pq > 0 q.e.d.


My solutions to c. and d. (slightly different than those of Bogdan) are:

c.

You must observe that:

[a₁+..+a_n]*[1/a₁+...+1/a_n] ≥ n+n[C]2*2=n+{n!/[(n-2)!*2!)]}*2=n^2

where I used:

{a_r/a_k}+[1/{a_r/a_k}] ≥ 2 (obviously (A+1/A) ≥ 2 when A > 0))

and n[C]2=combinations of n,2=n!/[(n-2)!*2!)]

d.

From:

[sum]_{r=1 to n} a_r = 1,

b. by replacing q with p/(p-1),

c. [sum] 1/a_r ≥ n² ---> [sum] a_r+1/a_r=[sum] a_r+[sum] 1/a_r ≥ 1+n²,

---> the required inequality.

[edited to change the layout of the demonstration]

 

[KLscilevothma]

Originally posted by Hurkyl
1/(1 - x) = &Sigma;_n=0..&infin; xⁿ

Differentiate both sides of the equation with respect to x
1/(1-x)² = &Sigma;_n=0..&infin;nx^n-1
sub x=1/2, then multiply both sides by 1/2
&Sigma;_n=0..&infin;n/n²

Take the 2nd derivatives on both sides of
1/(1 - x) = &Sigma;_n=0..&infin; xⁿ
2/(1-x)³=&Sigma;_n=0..&infin;(n²-n)x^n-2
sub x = 1/2 and multiply both sides by 1/4
The result follows.

edit: metacristi, I don't know inhomogeneous sequences

 

[KLscilevothma]

Thanks

Originally posted by bogdan
I think I got it...but it's quite simple...

if you know a)...then just put from 1/p+1/q=1, q=p/(p-1)...in d)...

sum[(Ar+1/Ar)^p]>n^(1-p)*[1+sum(1/Ar)]^p;
prove that sum(1/Ar)>n^2 this way...

sum(1/Ar)>n*(fact(1/Ar))^(1/n)=n*1/[fact(Ar)]^(1/n)>

> n/[sum(Ar)/n]=(n^2)/1...

Bogdan, I used the same method as you do (see my post above yours).

I now understand these 2 questions, thanks for your help.

Yes, it's simple. I knew how to do part a, b and c when I was at home and used about 15 minutes to finish them (exclude part d), but not in exam.. I was too nervous, I only had 20 minutes to think whether to choose this question or the other and finish it. I only had 1 day to revise for this algebra exam while I had put aside algebra for 6 months and studied calculus. (finding excuses to cover up my stupidity! )