# $\lim_{N\rightarrow \infty}\sum_{n=1}^N\frac{1}{N+n}=\log 2$ without using $\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}=\log2$

Is there a way of showing that $$\lim_{N\rightarrow\infty}\sum_{n=1}^N\frac{1}{N+n}=\int_1^2\frac{1}{x}dx=\log 2$$ without using $$\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}=\log2$$

I couldn't get rid of the small $n$ in the upper sum, so I tried rewriting the integral as a Riemann sum, but they still don't look anything alike. I'm preparing an exam btw.

Here's the rewritten integral: $\lim_{n\rightarrow\infty}\sum_{k=1}^n 2^{(k-1)/n}(2^{k/n}-2^{(k-1)/n})$ Doesn't seem to help with the sum though.

$$\frac1{N} \sum_{n=1}^N \frac1{1+\frac{n}{N}}$$
which in the limit as $N \to \infty$ is just the integral
$$\int_0^1 \frac{dx}{1+x} = \log{2}$$