Little-o of a summation

Question

I'm having trouble proving that

$$\sum_{k=1}^{n}\frac{1}{2k-1} = \log(n)/2 + \log(2) + γ/2 + o(1)$$ as $n → ∞$.

I honestly don't even know where to start, and I don't know what the $o(1)$ is supposed to signify. Can someone please help me?

Answer 1

Your expression looks like

$$\sum_{k=1}^{n}\frac{1}{2k-1} = \log(n)/2 + \log(2) + γ/2 + f(n)$$

with

$$f\in o(1) \Leftrightarrow \lim_{n\rightarrow \infty}\Big|\frac{f(n)}{1}\Big|=0$$

This is obviously equivalent to

$$\lim_{n\rightarrow \infty} f(n)=0$$

which means, that anything that is left after the summand $\frac{\gamma}{2}$, will vanish when considering the limit for $n\rightarrow\infty$