Simple Summation proof help! :) sum(n) i/(i^2) < 2
Phil IU Guy
Please!! Thanks!
-Phil
William Elliot
> How would you go about proving that sum(n) i/(i^2) < 2
>
Your notation is sloppy unto incoherence.
sum i/i^2 = sum 1/i which is know to diverge.
Doug Magnoli
coherently, you stand half a chance of getting an answer.
Do you mean sum(i=1,n) 1/i^2 ? If so, what's that i doing in the
numerator?
-Doug Magnoli
[Remove the two and the three for email.]
Alex.Lupas
> Hunh? What does sum(n) i/(i^2) mean? If you can write your question
> coherently, you stand half a chance of getting an answer.
> Do you mean sum(i=1,n) 1/i^2 ? If so, what's that i doing in the
> numerator?
> -Doug Magnoli
> > How would you go about proving that sum(n) i/(i^2) < 2
> Please!! Thanks!
Hi,
I try to solve your home work :
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Suppose that we know the inequalities (Cauchy) :
i) if 0<a<1 , -1<x =/= 0 , then (1+x)^{a} < 1+ax
ii) if a<0 or a>1 and -1< x =/= 0 , then (1+x)^{a} > 1+ax .
Consider 0<A<B . From above inequalities one finds
aB^{a-1}< (B^{a}-A^{a})/(B-A) < aA^{a-1} when 0<a<1 ,
(*) aA^{a-1}< (B^{a}-A^{a})/(B-A) < aB^{a-1} when a<0 or a>1 .
Denote S(n;s)=1/(1^{s})+ 1/(2^{s})+ ...+ 1/(n^{s})=
= Sum{k=1 to k=n} 1/(k^{s}) =
=\sum\limits_{k=1}^{n}\frac{1}{k^{s}}
and suppose s > 1 . Take a =1-s , a=k , b=k+1 in (*) .
One finds
(1-s)/(k^s) < 1/((k+1)^{s-1}) - 1/(k^{s-1}) < (1-s)/((k+1)^{s})
and (1-s<0 )
1/((k+1)^{s}) < ( 1/(1-s) )( 1/((k+1)^{s-1}) - 1/(k^{s-1}) )
{ k = 1,2,..., n-1}
By summing the above inequalities , we obtain
S(n;s) < 1+ ( 1/(s-1) ) ( 1-1/(n^{s-1}) )< 1+ 1/(s-1)
that is
===========================
(**) S(n;s) < s/(s-1) .
===========================
When s=2 you find form (**) your desired inequality S(n;2) <2 .
Remarks: I have tried to present proofs without Calculus.
However, (*) is a simple consequence of Lagrange mean theorem
and your inequality S(n;2) <2 may be proved in a easy way by
means of integrals.
( Of course, the above proof is not original.)There is an interesting book
by A.M. Yaglom and I.M.Yaglom (approx.Titel: Non-elementary problems solved
elementary ?? , tanslation from Russian) where you find many interesting
proofs.
Best regards, Alex.
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Alex.Lupas
Sorry,
I have solved another homework !
Here is the proof for Sum{i=1 to i=n}i/(2^i) (easier homework).
Let us denote S(n)=Sum{i=1 to i=n}i/(2^i) . Try to prove, for instance
by
induction, that
(1) S(k)= 2- (k+2)/(2^k) .
If you have performed the proof , then it is clear that S(n)< 2 .
Let us prove together this equality (1):
I denote by RHS(k) the right-side of (1).
STEP 1. (Verification of (1) for first positive integers)
k=1 S(1)=1/2 RHS(1)=1/2
k=2 S(2)=1 RHS(2)=1
STEP 2. (Assumption) Let us consider that (1) holds for k=1,2,...,n.
STEP 3. (Proof , namely (1)for k=n implies (1) for k=n+1). If (1) is
true for
k=n, then we may write
S(n+1)=S(n)+ (n+1)/(2^{n+1})= ( 2-(n+2)/(2^n) ) + (n+1)/(2^{n+1})=
= 2- (n+3)/(2^{n+1}) = RHS(n+1)
Because any positive integer has a succesor, it follows that (1) is
true for any (positive) natural number.
Another method to obtain (1) is by means of geometric progressions,
namely
it may be shown that
(1.1) Sum{i=1 to i=n}i*x^{i} = x( nx^{n+1}-(n+1)x^{n}+1)/((1-x)^2) .
Of course , (1.1) may be proved also by induction. Then put x=1/2 in
(1.1)
and you find (1).
Regards, Alex
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