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.
Show $\lim_{N\to \infty}\sum_{k=1}^{N}\frac{1}{k+N}=\ln(2)$
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0See [wikipedia](http://en.wikipedia.org/wiki/Euler–Mascheroni_constant) for $\gamma$. – 2012-07-02
4 Answers
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!
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0Please let me know how I can improve my answer. I really want to give you the best answer I can. – 2018-02-28
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 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} $
$f(x)=\lim_{n\to\infty} \sum \limits_{k=1}^n \frac{1}{k+\frac{n}{x}}$
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:
$f(x)=\frac{x}{1} -\frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}+ ....=\sum \limits_{k=1}^{\infty} (-1)^{k+1} \frac{x^k}{k}=\ln(x+1)$
$f(x)=\ln(x+1)$ $f(1)=\ln(2)$