Hola
Observa que:
\( \displaystyle\int_{1}^{n}\dfrac{1}{2x}dx\geq \displaystyle\sum_{k=2}^{n+1}{}\dfrac{1}{2k-1} \)
Por tanto:
\( \displaystyle\sum_{i=1}^n{\frac{1}{2i-1}}\leq \displaystyle\sum_{i=1}^{n+1}{\frac{1}{2i-1}}=1+\displaystyle\sum_{i=2}^{n+1}{\frac{1}{2i-1}}\leq \displaystyle\int_{1}^{n}\dfrac{1}{2x}dx+1 \)
Concluye...
Saludos.
