|
#1
|
|||
|
|||
interesting inequality
Show that 1+1/2+1/3+1/4+...1/p - lnp < 1 for all p >1
|
#2
|
|||
|
|||
Re: interesting inequality
[ QUOTE ]
Show that 1+1/2+1/3+1/4+...1/p - lnp < 1 for all p >1 [/ QUOTE ] 1+1/2+1/3+1/4+...1/p - lnp < 1 for all p >1, QED I challenge anyone to find a more parsimonious proof. |
#3
|
|||
|
|||
Re: interesting inequality
In particular, the 2^Kth partial sum of the harmonic series can be expressed as S = 1 + K/2. So in this case, we allow P = 2^K. Following, K = ln (P - 2).
Therefore, the partial sum can be expressed as S = 1 + [ln(P - 2)]/2. Then the entire expression can be rewritten as S - ln P < 1. This leads to: 1 + [ln(P - 2)]/2 -ln P < 1. [ln(P - 2)]/2 - ln P < 0 [ln(P - 2)]/2 < ln P ln(P - 2) < 2*ln P ln(P - 2) < ln (P^2) (P - 2) < P^2 P < P^2 + 2. Clearly, for P > 1, P^2 > P. So the statement 1 + 1/2 + 1/3 +...+ 1/P - ln P < 1 for all P > 1 holds. QED |
#4
|
|||
|
|||
Re: interesting inequality
Interesting Dano .
There is an alternative elementary way of solving this inequality . I'll let others think about it for a bit . |
#5
|
|||
|
|||
Re: interesting inequality
Even better one can show that
1+1/2+1/3+...+1/p -lnp <(p+1)/2p for all P>1 I improved on the last inequality . |
#6
|
|||
|
|||
Re: interesting inequality
The question turned out to be easier than I thought .
Lnp is just the area under the curve from 1 to p of 1/x. You can find an upper bound and a lower bound to express the approximate area as a summation . The sum of all the upper rectangles from [1,2] ,[2,3],..[p-1,p] is just 1+1/2+1/3+... 1/p The sum of all the lower rectangles is 1/2+1/3+...1/p . Lnp must be greater than the sum of all the lower rectangles and lower than the sum of all the upper rectangles . For the second problem , just use a trapezoid for the intervals [1,2],[2,3],[3,4]...[p-1,p] and show that lnp is less than the sum of all the trapezoids but greater than the sum of all the lower rectangles . |
#7
|
|||
|
|||
Re: interesting inequality
How about proving that 1+1/2+...+1/p- ln p actually has a limit as p goes to infinity? (For extra credit, prove that the limit is an irrational number...)
|
|
|