Show $\lim_{N\to \infty}\sum_{k=1}^{N}\frac{1}{k+N}=\ln(2)$ I have some difficulty to prove the following limit:
$$\lim_{N\to \infty}\sum_{k=1}^{N}\frac{1}{k+N}=\ln(2)$$
Can someone help me? Thanks.
 A: A common estimate for the Harmonic Numbers is
$$
\sum_{k=1}^n\frac1k=\log(n)+\gamma+O\left(\frac1n\right)\tag{1}
$$
where $\gamma$ is the Euler-Mascheroni constant.
Applying $(1)$, we get that
$$
\begin{align}
\sum_{k=1}^{N}\frac{1}{k+N}
&=\sum_{k=1}^{2N}\frac1k-\sum_{k=1}^N\frac1k\\
&=\left(\log(2N)+\gamma+O\left(\frac{1}{2N}\right)\right)-\left(\log(N)+\gamma+O\left(\frac1N\right)\right)\\
&=\log(2)+O\left(\frac1N\right)\tag{2}
\end{align}
$$
Taking the limit of $(2)$ as $N\to\infty$ yields
$$
\lim_{N\to\infty}\sum_{k=1}^{N}\frac{1}{k+N}=\log(2)\tag{3}
$$
A: $$f(x)=\lim_{n\to\infty} \sum \limits_{k=1}^n \frac{1}{k+\frac{n}{x}}$$
You are looking for $f(1)$ 
$$f(x)=\lim_{n\to\infty}\frac{x}{n} \sum \limits_{k=1}^n \frac{1}{1+\frac{kx}{n}}=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n (1-\frac{kx}{n}+\frac{k^2x^2}{n^2}-\frac{k^3x^3}{n^3}+....)=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n 1 -\lim_{n\to\infty} \frac{x^2}{n^2} \sum \limits_{k=1}^n k+\lim_{n\to\infty} \frac{x^3}{n^3} \sum \limits_{k=1}^n k^2-\lim_{n\to\infty} \frac{x^4}{n^4} \sum \limits_{k=1}^n k^3+.....$$
$$f(x)=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n 1 -\lim_{n\to\infty} \frac{x^2}{n^2} \sum \limits_{k=1}^n k+\lim_{n\to\infty} \frac{x^3}{n^3} \sum \limits_{k=1}^n k^2-\lim_{n\to\infty} \frac{x^4}{n^4} \sum \limits_{k=1}^n k^3+.....=\lim_{n\to\infty} \frac{x}{n} n -\lim_{n\to\infty} \frac{x^2}{n^2} (\frac{n^2}{2}+\frac{n}{2})+\lim_{n\to\infty} \frac{x^3}{n^3}   (\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6})-\lim_{n\to\infty} \frac{x^4}{n^4}(\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4})+....$$

$$\sum \limits_{k=1}^{n}   k^m=\frac{n^{m+1}}{m+1}+a_mn^m+....+a_1n=\frac{n^{m+1}}{m+1}+\sum \limits_{j=1}^m a_jn^j$$  where $a_j$ are constants.where aj are constants. More information about summation http://en.wikipedia.org/wiki/Summation

After solving limits. We get:
$$f(x)=\frac{x}{1} -\frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}+ ....=\sum \limits_{k=1}^{\infty} (-1)^{k+1} \frac{x^k}{k}=\ln(x+1)$$
$$f(x)=\ln(x+1)$$
$$f(1)=\ln(2)$$
A: Hint: It is a Riemann Sum that corresponds to the integral $$\int_{0}^1 \frac{1}{1+x} dx,$$ and this evaluates to $\ln 2$.
A: 
I just found this question and thought it might be instructive to present a way forward that relies on the Taylor series for $\log(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k-1}x^k}{k}$ from which we see that $\log(2)=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}$.  To that end, we proceed.


Note that we can write
$$\begin{align}
\sum_{k=1}^N \frac{1}{k+N}&=\sum_{k=N+1}^{2N}\frac1k\\\\
&=\sum_{k=1}^{2N}\frac1k-\sum_{k=1}^N\frac1k \tag 1
\end{align}$$
Next, we split the first sum on the right-hand side into sums of the even terms and odd terms and find that
$$\begin{align}
\color{blue}{\sum_{k=1}^{2N}\frac1k}-\color{red}{\sum_{k=1}^N\frac1k}&=\color{blue}{\left(\sum_{k=1}^N \frac{1}{2k}+\sum_{k=1}^N \frac{1}{2k-1}\right)}-\color{red}{\sum_{k=1}^N\frac1k}\\\\
&=\sum_{k=1}^N \frac{1}{2k-1}-\sum_{k=1}^N \frac{1}{2k}\\\\
&=\sum_{k=1}^{2N}\frac{(-1)^{k-1}}{k}\tag2
\end{align}$$
Letting $N\to \infty$ in $(2)$, we find
$$\begin{align}
\lim_{N\to \infty}\sum_{k=1}^N \frac{1}{k+N}&=\lim_{N\to \infty}\sum_{k=1}^{2N}\frac{(-1)^{k-1}}{k}\\\\
&=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\\\\
&=\log(2)
\end{align}$$
as was to be shown!
