Prove that $1<\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{3001}<\frac43$ 
Prove that $$1<\dfrac{1}{1001}+\dfrac{1}{1002}+\dfrac{1}{1003}+\dots+\dfrac{1}{3001}<\dfrac43 \, .$$

My work:
$$\begin{eqnarray*}
S&=&\bigg(\dfrac{1}{1001}+\dfrac{1}{3001}\bigg)+\bigg(\dfrac{1}{1002}+\dfrac{1}{3000}\bigg)+\ldots+\dfrac{1}{2001}\\
S&=&\dfrac{1}{4002}\bigg\{\bigg(\dfrac{1001+3001}{1001}+\dfrac{1001+3001}{3001}\bigg)+\ldots\bigg\}+\dfrac{1}{2001}\\
S&\ge& \dfrac{1}{4002}4\cdot1000+\dfrac{1}{2001}=1
\end{eqnarray*}$$
I could derive the left hand inequality with a $\ge$ sign though but could not do anything about the right hand inequality. Please help.
EDIT: I do not need to use the equality sign, I can rather use only strict inequality because there exist terms which are not equal so, the AM-HM inequality is actually a strict inequality here.
$$S> \dfrac{1}{4002}4\cdot1000+\dfrac{1}{2001}=1$$
Though I got the solution but I am looking for some non-calculus solutions too.
 A: For a solution without using calculus:  
For the upper bound, note for integers $1 \le k \le 1000$:
$$ \frac1{1000+k} + \frac1{3002-k} = \frac{4002}{(1000+k)(3002-k)}= \frac{4002}{4004001-(1001-k)^2} \le \frac{4002}{3004001}$$
where the equality is only when $k=1$.
$$\implies S = \sum_{k=1}^{1000} \left(\frac1{1000+k} + \frac1{3002-k} \right)+\frac1{2001} < \frac{4002}{3004001}\cdot 1000+\frac1{2001} = \frac{8011006001}{6011006001} < \frac43$$
For the lower bound here is another way, using $AM > HM$ for distinct numbers, we get
$$\frac1{1000+k} + \frac1{3002-k}> \frac2{2001} \qquad \text{for }k = 1, 2, 3, \dots 1000$$
$$\implies S = \sum_{k=1}^{1000} \left(\frac1{1000+k} + \frac1{3002-k} \right)+\frac1{2001} > \frac{2}{2001} \cdot 1000+\frac1{2001}=1$$
A: $$\log{k+1\over k}=\int_k^{k+1}{dx\over x}<{1\over k}<\int_{k-1}^k{dx\over x}=\log{k\over k-1}\qquad(k>1)\ .$$
Summing over $k$ produces telescoping series on the left hand and the right hand side. Doing the computations one finds
$$\log{3002\over1001}<S<\log{3001\over 1000}\ .$$
Here the left hand side is obviously $>1$. For the right hand side we note that
$$e^{4\over3}>1+{4\over3}+{(4/3)^2\over2}={29\over9}>{3001\over1000}\ ,$$ and this shows that
$$\log{3001\over 1000}<{4\over3}\ .$$
Update: The OP has desired a calculus-free approach. For the upper estimate one could argue as follows: Using the splitting
$$S=\sum_{k=1001}^{1350}{1\over k}+\sum_{k=1351}^{1800}{1\over k}+\sum_{k=1801}^{2400}{1\over k}+\sum_{k=2401}^{3000}{1\over k}+{1\over3001}$$
one obtains 
$$S<{350\over1000}+{450\over1350}+{600\over1800}+{600\over2400}+{1\over3001}={76\over60}+{1\over3001}<{4\over3}\ .$$
A: Let $k$ be a fixed integer greater than $0$. 
It holds that $\frac{1}{1000+k}\leq \frac{1}{1000+x}$for any $x \in [k-1,k]$.
Integrating for fixed $k \in [1,2001]$ yields $$\frac{1}{1000+k}\leq \int_{k-1}^{k}\frac{1}{1000+x}dx$$
Hence
$$\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\ldots+\frac{1}{3001} \leq \int_{0}^{2001}\frac{1}{1000+x}dx=\ln\left(\frac{3001}{1000}\right)$$
And $$\ln\left(\frac{3001}{1000}\right)\sim 1.1 < \frac{4}{3}$$
A: For $k=0,1 \cdots 2000$ we have
$$ \frac{1}{1001+k} < \frac{1}{1000}\left(1 - \frac{k}{3000}\right)$$
Explanation: let the left side be $f(k)$ and the right side $g(k)$. Then $f(0)=1/1001$, $g(0) = 1/1000$; while $f(2000)=1/3001$ and  $g(2000) =1/3000$.
Hence $f < g$ in the extreme points - and because $f$ is convex, it must also be below $g$ in the intermediate values, because $g$ is linear.
Then $$\sum_{k=0}^{2000} \frac{1}{1001+k} < \frac{1333}{1000} <\frac{4}{3}$$
A: For the lower bound we can straight away use Titu's Lemma:
Let $$M=\sum_{k=1}^{2001} \frac{1}{1000+k}$$
By Generalized Titu's Lemma and Arithmetic Progression we get
$$
\begin{aligned}
 & M>\frac{(1+1+1+\cdots+1)^2}{1001+1002+1003+\ldots+3001} \\
 \\
 & \Rightarrow M>\frac{2001^2}{\left(\frac{2001}{2} \times 4002\right)}=1
\end{aligned}
$$
