Show that $\lim_{n \to \infty} \sum_{k=3}^n \frac{2k}{k^2+n^2+1} = \ln(2)$

Question

Show that

$$\lim_{n \to \infty} \sum_{k=3}^n \frac{2k}{k^2+n^2+1} = \ln(2)$$

How many ways are there to prove it ?

Is there a standard way ?

I was thinking about making it a Riemann sum. Or telescoping.

What is the easiest way ? What is the shortest way ?

Answer 1

Observe \begin{align} \sum^n_{k=3}\frac{2k}{n^2+k^2+1} = \frac{1}{n}\sum^n_{k=3} \frac{2(k/n)}{1+n^{-2}+(k/n)^2} \end{align} then we have \begin{align} \frac{1}{1+n}\sum^n_{k=3} \frac{2k/(1+n)}{1+k^2(1+n)^{-2}} \leq \frac{1}{n}\sum^n_{k=3} \frac{2(k/n)}{1+n^{-2}+k^2n^{-2}} \leq \frac{1}{n}\sum^n_{k=3} \frac{2(k/n)}{1+k^2n^{-2}}. \end{align} Hence it follows \begin{align} \int^1_0 \frac{2x}{1+x^2} \ dx=\lim_{n\rightarrow \infty}\frac{1}{1+n}\sum^n_{k=3} \frac{2k/(1+n)}{1+k^2(1+n)^{-2}} \leq \lim_{n\rightarrow \infty}\frac{1}{n}\sum^n_{k=3} \frac{2(k/n)}{1+n^{-2}+k^2n^{-2}} \leq \lim_{n\rightarrow \infty}\frac{1}{n}\sum^n_{k=3} \frac{2(k/n)}{1+k^2n^{-2}} = \int^1_0 \frac{2x}{1+x^2}\ dx \end{align} which means \begin{align} \lim_{n\rightarrow \infty}\sum^n_{k=3} \frac{2k}{1+k^2+n^2} = \int^1_0 \frac{2x}{1+x^2}\ dx = \log 2. \end{align}

Answer 2

Note that

$$\sum_{k=3}^n \frac{2k}{k^2+n^2+1} = \sum_{k=1}^n \frac{2k}{k^2+n^2+1} - \frac{2}{2+n^2} - \frac{4}{5+n^2}.$$

We can ignore the last two terms since they converge to $0$.

Consider

$$\sum_{k=1}^n \frac{2k}{k^2+n^2+1} = \frac{1}{n}\sum_{k=1}^n \frac{2(k/n)}{1+(k/n)^2 + (1/n^2)}. $$

This is almost a Riemann sum for $\int_0^1 2x/(1 +x^2) \, dx$ except for the annoying term $1/n^2$.

However, it is valid to take a limit of a double sequence

$$S_{mn} =\frac{1}{n}\sum_{k=1}^n \frac{2(k/n)}{1+(k/n)^2 + (1/m^2)}, $$

as

$$\lim_{n \to \infty}S_{nn} = \lim_{n \to \infty} \lim_{m \to \infty} S_{mn},$$

since the inner limit on the RHS exhibits uniform convergence.

Thus,

$$\lim_{n \to \infty}\sum_{k=3}^n \frac{2k}{k^2+n^2+1} = \lim_{n \to \infty} \lim_{m \to \infty}\frac{1}{n}\sum_{k=1}^n \frac{2(k/n)}{1+(k/n)^2 + (1/m^2)} \\ = \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n \frac{2(k/n)}{1+(k/n)^2} \\ = \int_0^1 \frac{2x}{1+ x^2} \, dx \\ = \log 2$$

Answer 3

Just added for your curiosity.

Riemann sum is certainly the fastest way to do it but you can also do it differently using generalized harmonic numbers (after partial fraction decomposition) and obtain $$S_n=\sum_{k=3}^n \frac{2k}{k^2+n^2+1}=-H_{2-\sqrt{-n^2-1}}+H_{n-\sqrt{-n^2-1}}-H_{\sqrt{-n^2-1}+2}+H_{n+\sqrt{-n^2-1}}$$ Now, using the asymptotics $$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{n^4}\right)$$ you should arrive to $$S_n=(\log (1-i)+\log (1+i))+\frac{1}{2 n}-\frac{20}{3 n^2}-\frac{1}{4 n^3}+O\left(\frac{1}{n^4}\right)$$ that is to say $$S_n=\log (2)+\frac{1}{2 n}-\frac{20}{3 n^2}-\frac{1}{4 n^3}+O\left(\frac{1}{n^4}\right)$$ which, for sure, shows the limit but also how it is approached; moreover, it gives a very good approxiamtion of $S_n$.

For example $S_{10}=\frac{1402864984}{2067340275}\approx 0.6786$ while the above expansion would give $\log (2)-\frac{203}{12000}\approx 0.6762$.

Edit for clarity

Let $\alpha=-i\sqrt{n^2+1}$, $\beta=i\sqrt{n^2+1}$ be the roots of $k^2+n^2+1=0$. Partial fraction decomposition leads to $$\frac{2k}{k^2+n^2+1}=\frac{2}{\alpha -\beta }\left(\frac \alpha {k-\alpha}-\frac \beta {k-\beta} \right)$$ Now, using $$\sum_{k=3}^n\frac 1{k-a}=\psi ^{(0)}(-a+n+1)-\psi ^{(0)}(3-a)=H_{n-a}-H_{2-a}$$