How to find the value $\frac {1}{1001}$+$\frac {1}{1002}$+$ \frac {1}{1003}$. . . . $\frac {1}{3001}$ How to find the value of $X$?
If $X$= $\frac {1}{1001}$+$\frac {1}{1002}$+$
\frac {1}{1003}$. . . . $\frac {1}{3001}$
 A: 
How to find the value of X ?

You don't. All you can do is to approximate it with $\ln3000-\ln1000=\ln\dfrac{3000}{1000}=\ln3$.
A: Mathematica gives an approximation of
1.098612251
Ref.: http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427egmfbkbu8jv&mail=1
Just to note that $\ln 3 \approx 1.098612288$.  So the approximation given before are VERY good...
A: The exact answer is $$\begin{align}
H_{3001}-H_{1000}&=(\gamma+\psi_0(3002))-(\gamma+\psi_0(1001))\\ \,\\&=\dfrac{\Gamma'(3002)\Gamma(1001)-\Gamma'(1001)\Gamma(3002)}{\Gamma(1001)\Gamma(3002)}\\\,\,\\
&\approx 1.09861225...
\end{align}$$ where $H_n$ is the $n$-th Harmonic number defined as: $$H_n:=\sum_{k=1}^n \dfrac1k$$ which can be expressed analytically by the formula: $H_n=\gamma+\psi_0(n+1)$ where $\gamma$ is the Euler-Mascheroni constant and $\psi_0$ is the Digamma function defined as: $$\psi_0(n):=\dfrac{\mathrm d}{\mathrm dn}\ln \big(\Gamma (n)\big)=\dfrac{\Gamma'(n)}{\Gamma(n)}$$ where $\Gamma(n)$ is the gamma function which is equal to $\Gamma(n)=(n-1)!$.
