How do I prove this Euler-Mascheroni limit?

Question

There is a number called the Euler-Mascheroni constant that is defined as the limiting difference between the harmonic series and the natural logarithm. In other words: $$ \gamma = \lim_{N \to \infty}\,\,\left[-\ln\left(N\right) + \sum_{k = 1}^{N}\frac{1}{k}\right] $$ On desmos, I shifted $\ln$ by a number $a$ and it still converges to $\gamma$. This means that, for all numbers $a$: $$ \gamma = \lim_{N \to \infty}\,\,\left[-\ln\left(N + a\right) + \sum_{k = 1}^{N}\frac{1}{k}\right] $$ How do I prove that this is true $?$. I think it has to do with the concavity of the logarithm but I don't know how to use this.

Answer 1

$$ \begin{align*} &\lim_{N\rightarrow\infty} \left(-\ln(N+a) + \sum_{k=1}^N \frac{1}{k}\right) \\ &= \lim_{N\rightarrow\infty} (\ln(N) - \ln(N+a)) + \lim_{N\rightarrow\infty} \left(-\ln(N) + \sum_{k=1}^N \frac{1}{k}\right) \\ &= \gamma + \lim_{N\rightarrow\infty} \ln\left(\frac{N}{N+a}\right) \\ &= \gamma + \lim_{N\rightarrow\infty} \ln\left(1-\frac{a}{N+a}\right), \hspace{1em}\text{ since } \frac{N}{N+a} = \frac{N+a-a}{N+a}\\ &= \gamma + \ln(1), \hspace{8em}\text{ since ln is continuous and } \frac{1}{N+a} \rightarrow 0\\ &= \gamma \end{align*} $$