Mathematica tells me that $\sum\limits_{i=1}^n \frac1{2i-1}$ is equal to $\frac12 H_{n-1/2}+\log\,2$, where $H_n$ is a harmonic number.

Why is this true? Is there a general strategy for solving sums of the form $\sum\limits_{i=1}^n \frac1{ai+b}$?

Mathematica tells me that $\sum\limits_{i=1}^n \frac1{2i-1}$ is equal to $\frac12 H_{n-1/2}+\log\,2$, where $H_n$ is a harmonic number.

Why is this true? Is there a general strategy for evaluating sums of the form $\sum\limits_{i=1}^n \frac1{ai+b}$?

Mathematica tells me that $\sum _{i=1}^n \frac{1}{2 i-1}$ is equal to $\frac{1}{2} \text{HarmonicNumber}\left[n-1/2\right]+\text{Log}[2]$.

Why is this true? Is there a general strategy for solving sums of the form $\sum _{i=1}^n \frac{1}{a i + b}$?

Mathematica tells me that $\sum\limits_{i=1}^n \frac1{2i-1}$ is equal to $\frac12 H_{n-1/2}+\log\,2$, where $H_n$ is a harmonic number.

Why is this true? Is there a general strategy for solving sums of the form $\sum\limits_{i=1}^n \frac1{ai+b}$?