interesting inequality

[jay_shark]

Show that 1+1/2+1/3+1/4+...1/p - lnp < 1 for all p >1

[arahant]

[ 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.

[Dan.]

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

[jay_shark]

Interesting Dano .

There is an alternative elementary way of solving this inequality . I'll let others think about it for a bit .

[jay_shark]

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 .

[jay_shark]

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 .

[astvald]

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...)