Prove that \[ \lim_{n\to\infty} \left( \frac1{n+1}+\frac1{n+2}+\cdots+\frac1{2n} \right) =\log2. \]
Solution
1. For every \(n\geq1\),\[ \sum_{k=n+1}^{2n}\frac1k = \frac1n \sum_{j=1}^{n} \frac{1}{1+j/n}. \]
Put \(k=n+j\). Then \(k\) runs from \(n+1\) through \(2n\) exactly when \(j\) runs from \(1\) through \(n\), and \[ \frac1{n+j} = \frac1n\frac1{1+j/n}. \] Summing gives the identity.
2. The right-hand side in step 1 converges to\[ \int_0^1\frac{dx}{1+x}. \]
The function \[ g(x)=\frac1{1+x} \] is continuous on \([0,1]\). The expression \[ \frac1n \sum_{j=1}^{n}g\left(\frac jn\right) \] is its right-endpoint Riemann sum for the uniform partition of \([0,1]\) into \(n\) subintervals. Hence it converges to the displayed integral.
3. One has\[ \int_0^1\frac{dx}{1+x}=\log2. \]
Direct integration gives \[ \int_0^1\frac{dx}{1+x} = \left[\log(1+x)\right]_0^1 = \log2. \]
4. Therefore\[ \boxed{ \lim_{n\to\infty} \left( \frac1{n+1}+\frac1{n+2}+\cdots+\frac1{2n} \right) = \log2. } \]
Combine steps <1>1–<1>3.
5. Q.E.D.
Step 4 is the required conclusion.