Question

Prove:\[1 < \frac{1}{1001} + \frac{1}{1002} + \frac{1}{1003} + \cdots + \frac{1}{3001} < 1 \frac{1}3\]

Answer 1

The left inequality is a standard AM-HM one. How about the right?

Answer 2

@ganeshie8

Answer 3

Calc seems to work. Algebra?

Answer 4

Since the required sum is only about 1.0986 with a big margin from 4/3, we can brute force the right inequality:
Sum < 500(1/1001+1/1501+1/2001+1/2501)=1.3263<4/3

Answer 5

Here's a more refined proof for the second inequality:
We define \(s(m)=\frac{1}{1001+m}+\frac{1}{3001-m} \)
and it can be shown that \(s(i)>s(i+1) \) for \(0\le i\le 999\).........(1)
\(S=\sum_{k=1001}^{k=3001}1/k\)
\(=(\sum_{m=0}^{m=1000}s(m) ~) ~~-1/2001\)
\(< 1001s(0)-1/2001 \)...............using (1)
\(= 1.33305573 \)
\(< 4/3\)


Proof for (1):
\(s(n)-s(n+1))=-1/(n+1002)+1/(n+1001)+1/(3001-n)-1/(3000-n)\)
\(=1/(n^2+2003n+1003002)-1/(n^2-6001n+9003000)\)
\(>0~\forall ~0\le n\le 999\)