$\lim_{{n \to \infty}} \sum_{{k=1}}^{n} \arctan\left(\frac{1}{k}\right) - \ln n$

Question

$\arctan(x) = \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1}x^{2m+1}$

\begin{align*} \arctan\left(\frac{1}{k}\right) &= \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1}\left(\frac{1}{k^{2m+1}}\right) &= \frac{1}{k} + \sum_{m=1}^{\infty} \frac{(-1)^m}{2m+1}\left(\frac{1}{k^{2m+1}}\right) \end{align*} We have that \begin{align*} \sum_{k=1}^{\infty} \frac{1}{k^{2m+1}} &= \zeta(2m+1)\\ \end{align*}

converges, provided that $m>0$ and

$\lim_{{n \to \infty}} \left( \sum_{{k=1}}^{n} \frac{1}{k} - \ln(n) \right) = \gamma$

where $\gamma$ is the Euler-Mascheroni constant.

Need help in further

Answer 1

have

$\arctan(x)=\sum_{m=0}^{\infty} \dfrac{(-1)^m}{2m+1} x^{2m+1}$

And as such we find

$\arctan\left(\frac{1}{k}\right)=\sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1} \frac{1}{k^{2m+1}}=\frac{1}{k} + \sum_{m=1}^{\infty} \frac{(-1)^m}{2m+1} \frac{1}{k^{2m+1}}$

We have that

$\displaystyle \sum_{k=1}^{\infty} \dfrac{1}{k^{2m+1}}=\zeta(2m+1)$

converges, provided that $m>0$, and

$\lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{k}-\ln(n)=\gamma$

where $\gamma$ is the Euler-Mascheroni constant.

We find

$\lim_{n\to \infty} \left(\sum_{k=1}^n \arctan\left(\frac{1}{k}\right)-\ln(n)\right)=\gamma+\sum_{m=1}^{\infty} \dfrac{(-1)^m}{2m+1}\zeta(2m+1)$

Note that the alternating series at the right does converge, since its terms tend to $0$, so the swap of the limit and the series (and the swap of the sums) is allowed. Whether this can be rewritten in closed form, that I don’t know, you might want to search for zeta values at the odd positive integers, but I don’t think alternative expressions allow for an easier attack.

Addendum

A more concise answer can be obtained by taking a detour to the complex numbers. Lets rewrite the arctangent to its complex form, by using the following sketch

If we associate the point $(k,1)$ with the complex number $k+i$and the point $(k,-1)$ with the complex number $k-i$ we obtain

$\ln(k+i)=a+i \text{Arg}(k+i)=a+i \arctan(1/k)\tag*{}$ [math]\ln(k-i)=a-i\text{Arg}(k-i)=a- i \arctan(1/k)\tag*{}$

where the real part of both expressions $a$ is equal to the logarithm of the distance of these points to the origin. Subtract both equations and the real parts drop out

$2i \arctan(1/k)=\ln(k+i)-\ln(k-i)\tag*{}$

or

$\arctan(1/k)=\dfrac{1}{2i} (\ln(k+i)-\ln(k-i))\tag*{}$

We may thus replace the given sum as follows

$\dfrac{1}{2i} \sum_{k=1}^n \ln(k+i)-\ln(k-i)\tag*{}$

which can be rewritten to

$\dfrac{1}{2i} \ln\left(\dfrac{\prod_{k=1}^n k+i}{\prod_{k=1}^n k-i}\right)=\dfrac{1}{2i}$ $\ln\left(\dfrac{(1+i)^{\underline{n}}}{(1-i)^{\underline{n}}}\right)\tag*{}$

where the last notation indicates the rising factorial (defined through the correspondig terms at the left).

We still have to incorporate the term $-\ln(n)$, which we may achieve by including

$-\dfrac{1}{2i}\ln(n)=-\dfrac{1}{2i} (2i \ln(n))=\dfrac{1}{2i} \ln\left(n^{-2i}\right)\tag*{}$

A slight warning is in order here. The logarithm is to measure the angle of $n^{2i}$ with respect to the real axis, including all extra rotations! For instance if you ask Wolfram Alpha to compute $\ln(1000^{-2i})$ it will respond with the principal branch of the logarithm and reduce the argument modulo $2\pi$. That’s not what is written above!

All in all we find that the expression reduces to

$\dfrac{1}{2i} \ln\left(n^{-2i}\dfrac{\prod_{k=1}^n k+i}{\prod_{k=1}^n k-i}\right)\tag*{}$

and we are to compute

$\lim_{n \to \infty} \dfrac{1}{2i} \ln\left(n^{-2i}\dfrac{(1+i)^{\underline{n}}}{(1-i)^{\underline{n}}}\right)\tag{1}$

Time to let Gauss enter the stage, who started his investigation of $\Gamma(z)$ by defining

$\Gamma(z)$=$\lim_{n\to \infty} \dfrac{n^z n!}{(z)^{\underline{n+1}}}\tag*{}$

From which we obtain

$\dfrac{\Gamma(z_1)}{\Gamma(z_2)}=\lim_{n \to \infty} \dfrac{n^{z_1}}{n^{z_2}} \dfrac{(z_2)^{\underline{n+1}}}{(z_1)^{\underline{n+1}}}=\lim_{n \to \infty} \dfrac{(n-1)^{z_1}}{(n-1)^{z_2}} \dfrac{(z_2)^{\underline{n}}}{(z_1)^{\underline{n}}}\tag*{}$

We also have

$\dfrac{(n-1)^{z_1}}{(n-1)^{z_2}}=e^{(z_1-z_2)\ln(n-1)}=(n-1)^{z_1-z_2}\tag*{}$

With the same annotation as previously. Compare with $(1)$ and pick $z_1=1-i, z_2=1+i$

$\dfrac{\Gamma(1-i)}{\Gamma(1+i)}=\lim_{n \to \infty} \dfrac{(n-1)^{-2i}(1+i)^{\underline{n}}}{(1-i)^{\underline{n}}}\tag{2}$

Finally observe that

$\lim_{n \to \infty} \dfrac{(n-1)^{-2i}}{n^{-2i}}=1\tag*{}$

which is a consequence of

$\lim_{n \to \infty} \ln(n-1)-\ln(n)=\lim_{n\to \infty} \ln(1–1/n)=0\tag*{}$

and we find

$\lim_{n\to \infty} \left(\sum_{k=1}^n \arctan\left(\frac{1}{k}\right)-\ln(n)\right)=\dfrac{1}{2i} \ln\left(\dfrac{\Gamma(1-i)}{\Gamma(1+i)}\right)=\dfrac{i}{2}\ln\left(\dfrac{\Gamma(1+i)}{\Gamma(1-i)}\right)\tag*{}$

Answer 2

I wouldn't recommend solving double series with rearrangements from a power series to a zeta function (i.e., $\displaystyle \sum_k\sum_i k^i=\sum_i\sum_k k^i$), since it's usually the former that is easier to solve (e.g., by a geometric series) while the latter makes things even more difficult.

Instead, since you have a summand that has convenient differential properties, where the trouble is caused by the summation over discrete integers, you can ameliorate that by smoothing out the sum, converting it to an integral. Hence, try the Euler-Maclaurin formula. This gives you $$\sum_{k=1}^{x}\arctan\frac{1}{k} =\int_{1}^{x}\arctan\frac{1}{t}dt+\frac{\arctan1+\arctan\frac{1}{x}}{2}+\sum \frac{B_{2k}}{(2k)!}\ldots \\=\frac{1}{2}\left(-\frac{\pi}{4}-\ln2\right)+\frac{\ln\left(x^{2}+1\right)}{2}+\left(x+\frac{1}{2}\right)\arctan\frac{1}{x}+\sum\ldots$$

The middle terms dependent on $x$ tend to $\ln x +1$, so in the limiting difference they become $+1$. The remainder Bernoulli number terms do annoyingly retain some constant order parts, but from a brief numerical evaluation, those seem to vanish after the 4th term. We could evaluate that more thoroughly with the expression for the derivatives of arctan given here that would give us a power series since the arctan derivatives become rational functions, but I'll leave that to you. Ultimately, numerical evidence suggests the constant term should be $0.302$, which is a little more than the $0.207$ extracted in the initial terms.