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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.

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Could you add some description about the rôle of $\Psi$? –  Siminore Jul 2 '12 at 11:33
I know the analytic expression of this sum is: $\Psi(2N+1)-\Psi(N+1)$ –  Riccardo.Alestra Jul 2 '12 at 11:35
Do you know that $\sum_1^n(1/m)=\log n+\gamma+o(1)$? –  Gerry Myerson Jul 2 '12 at 11:54
See wikipedia for $\gamma$. –  Frank Science Jul 2 '12 at 13:25

3 Answers 3

up vote 8 down vote accepted

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 $$ \lim_{N\to\infty}\sum_{k=1}^{N}\frac{1}{k+N}=\log(2)\tag{3} $$

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Hint: It is a Riemann Sum that corresponds to the integral $$\int_{0}^1 \frac{1}{1+x} dx,$$ and this evaluates to $\ln 2$.

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$$f(x)=\lim_{n\to\infty} \sum \limits_{k=1}^n \frac{1}{k+\frac{n}{x}}$$

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:

$$f(x)=\frac{x}{1} -\frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}+ ....=\sum \limits_{k=1}^{\infty} (-1)^{k+1} \frac{x^k}{k}=\ln(x+1)$$

$$f(x)=\ln(x+1)$$ $$f(1)=\ln(2)$$

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