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Asymptotics: Lets look at the quantity

$$S(n)=(-1)^{n}(n+1)\binom{i}{n+1}=i\prod_{k=1}^{n+1}\left(1-\frac{i}{k}\right).$$ It's just your binomial coefficient above with the $(-1)^{n+1}$ factored in, and an extra $n+1$ so it factors nicely as a product.

Claim: We have that

$$S(n)=\sqrt{\frac{\sinh{\pi}}{\pi}}e^{iC_{0}}e^{-i\log n}\left(1+O\left(\frac{1}{n}\right)\right),$$ where

$$C_{0}=\frac{-\pi}{2}-1+\int_0^\infty \frac{\{x\}}{1+x^2}dx +O\left(\frac{1}{n}\right) =\arg(\Gamma(i))\approx -1.872.$$

In particular, the angle moves around the circle like $\log n$.

Application to your question: The above claim shows that $$f_{n+1} = (-1)^{n+1} n! \sqrt{\frac{\sinh{\pi}}{\pi}}\cos(-\log n+C_0)\left(1+O\left(\frac{1}{n}\right)\right)$$ and $$g_{n+1} \sim (-1)^{n+1} n! \sqrt{\frac{\sinh{\pi}}{\pi}} \sin(-\log n+C_0)\left(1+O\left(\frac{1}{n}\right)\right).$$

In particular, the ratio $g_n/f_n$ can be made arbitrarily large or small.

Proof of the claim: We first note that the size is

$$\sqrt{\prod_{k=1}^{n}\left(1+\frac{1}{k^{2}}\right)}=\sqrt{\prod_{k=1}^{\infty}\left(1+\frac{1}{k^{2}}\right)}+O\left(\frac{1}{n}\right).$$ To evaluate this product, recall the Weierstrass product for the Gamma function $$\left(\Gamma(z)\right)^{-1}=ze^{\gamma z}\prod_{k=1}^{\infty}\left(1+\frac{z}{k^{2}}\right)e^{-\frac{z}{r}}.$$ From this it follows that $$\frac{1}{|\Gamma(i)|^{2}}=\frac{1}{\Gamma(i)\Gamma(-i)}=\prod_{k=1}^{\infty}\left(1+\frac{1}{k^{2}}\right).$$ Using the identity $$\Gamma(x)\Gamma(-x)=-\frac{\pi}{x\sin\left(\pi x\right)},$$ we now have that $$\frac{1}{\Gamma(i)\Gamma(-i)}=\frac{-i\sin(i\pi)}{\pi}=\frac{\sinh(\pi)}{\pi},$$ which gives rise to the $\sqrt{\frac{\sinh(\pi)}{\pi}}$ term. Moving on to the evaluation of the angle, by looking at each triangle, and noting that the argument is additive when multiplied, we get that the argument equals

$$-\sum_{k=1}^{n}\tan^{-1}\left(\frac{1}{k}\right).$$

The negative sign arises since we are working in the fourth quadrant. By looking at the Taylor series for $\tan^{-1}$ we see that the above is $\log n+O(1)$, however, I would like to compute this argument more precisely, and obtain the constant. Lets compare our $\tan^{-1}$ series to the harmonic series. Rewriting things in terms of a Riemann Stieltjes integral, and using summation by parts, we have that

$$\sum_{k=1}^{n}\tan^{-1}\left(\frac{1}{k}\right)=\int_{0}^{n}\tan^{-1}\left(\frac{1}{x}\right)d\left[x\right]=[n]\tan^{-1}(1/n)\int_{0}^{n}\frac{\left[x\right]}{1+x^{2}}dx.$$

Pulling out the main term with the identity $[x]=x-\{x\}$, the above equals

$$\int_{0}^{n}\frac{x}{1+x^{2}}dx-\int_{0}^{n}\frac{\{x\}}{1+x^{2}}dx.$$

Since the first integral evaluates to $\frac{1}{2}\log(1+x^2)$, we have that $$\sum_{k=1}^{n}\tan^{-1}\left(\frac{1}{k}\right)=\log n +1-\int_0^\infty \frac{\{x\}}{1+x^2}dx +O\left(\frac{1}{n}\right).$$

Acknowledgements: I would like to thank Noam Elkies for pointing out that $$\prod_{k=1}^\infty \sqrt{1+\frac{1}{k^2}}=\frac{1}{|\Gamma(i)|}=\sqrt{\frac{\sinh(\pi)}{\pi}}$$ in the comments.

Edit: Be careful when using Wolfram Alpha for series which come your way. Sometimes it is wrong. Example: The series $\sum_{k=1}^\infty \frac{1}{k(k^2+u^2)}$, and Fixed the series constants appearing. Interestingly $\sum_{k=1}^\infty$\Gamma(i)=\sqrt{\frac{\pi}{\sinh{\pi}}}\exp\left(i\left(\frac{-\pi}{2}-1+\int_0^\infty \frac{1}{k^2(k^2+u^2)}$both evaluate to $$\frac{1}{2z^4}-\frac{\pi frac{\{x\}}{1+x^2}dx \coth(\pi z)}{2z^3}+\frac{\pi^2}{6z^2}$$ which is impossible. The latter is correct, however the former is not. This led to a small error in my previous constant.right)\right).$$9 added 20 characters in body; deleted 6 characters in body; added 16 characters in body; edited body $$S(n)=(-1)^{n}(n+1)\binom{i}{n+1}=i\prod_{k=1}^{n}\left(1-\frac{i}{k}\right).$$S(n)=(-1)^{n}(n+1)\binom{i}{n+1}=i\prod_{k=1}^{n+1}\left(1-\frac{i}{k}\right).$$ It's just your binomial coefficient above with the$(-1)^n$(-1)^{n+1}$ factored in, and an extra $n+1$ so it factors nicely as a product.

$$S(n)=\sqrt{\frac{\sinh{\pi}}{\pi}}e^{i(\frac{\pi}{4}-C_{0})}e^{-i\log S(n)=\sqrt{\frac{\sinh{\pi}}{\pi}}e^{iC_{0}}e^{-i\log n}\left(1+O\left(\frac{1}{n}\right)\right),$$

$$C_{0}=\gamma-\frac{\pi^{2}}{6}+\int_{0}^{1}\left(\frac{\pi\coth(\pi u)}{2u}-\frac{1}{2u^{2}}\right)du\approx0.33815.$$ $C_{0}=\frac{-\pi}{2}-1+\int_0^\infty \frac{\{x\}}{1+x^2}dx +O\left(\frac{1}{n}\right) =\arg(\Gamma(i))\approx -1.872.$$$$f_{n+1} = (-1)^n -1)^{n+1} n! \sqrt{\frac{\sinh{\pi}}{\pi}}\cos(\log n+C_0+\pi/4)\left(1+O\left(\frac{1}{n}\right)\right)$$sqrt{\frac{\sinh{\pi}}{\pi}}\cos(-\log n+C_0)\left(1+O\left(\frac{1}{n}\right)\right)$$ and $$g_{n+1} \sim (-1)^n -1)^{n+1} n! \sqrt{\frac{\sinh{\pi}}{\pi}} \sin(\log n+C_0+\pi/4)\left(1+O\left(\frac{1}{n}\right)\right).$$sin(-\log n+C_0)\left(1+O\left(\frac{1}{n}\right)\right).$$ The negative sign arises since we are working in the fourth quadrant. By looking at the Taylor series for \tan^{-1} we see that the above is \log n+O(1), however, I would like to compute this argument more precisely, and obtain the constant. Lets compare our \tan^{-1} series to the harmonic series. Notice that$$\sum_{k=1}^{n}\tan^{-1}\left(\frac{1}{k}\right)-\frac{1}{k}=-\sum_{k=1}^{n}\int_{0}^{\frac{1}{k}}\frac{x^{2}}{1+x^{2}}dx,$$Since \tan^{-1}(x)=\int_0^x \frac{1}{1+x^2}dx. The above then equals$$\sum_{k=1}^{n}\frac{1}{k}\int_{0}^{1}\frac{u^{2}}{k^{2}+u^{2}}du=\int_{0}^{1}\sum_{k=1}^{n}\frac{1}{k}\frac{u^{2}}{k^{2}+u^{2}}du.$$For |u|<1, Rewriting things in terms of a Riemann Stieltjes integral, and using summation by parts, we have that$$\sum_{k=1}^{n}\frac{1}{k}\frac{u^{2}}{k^{2}+u^{2}}=\sum_{k=1}^{\infty}\frac{1}{k}\frac{u^{2}}{k^{2}+u^{2}}+O\left(\frac{1}{n^{2}}\right),$$and we may now use \sum_{k=1}^{n}\tan^{-1}\left(\frac{1}{k}\right)=\int_{0}^{n}\tan^{-1}\left(\frac{1}{x}\right)d\left[x\right]=[n]\tan^{-1}(1/n)\int_{0}^{n}\frac{\left[x\right]}{1+x^{2}}dx.$$ Pulling out the fact that this infinite series is related to main term with the cot function function. Indeed,identity $$\sum_{k=1}^{\infty}\frac{1}{k}\frac{u^{2}}{k^{2}+u^{2}}=\frac{1}{2u^{2}}-\frac{\pi\coth(\pi u)}{2u}+\frac{\pi^{2}}{6},$$ and so[x]=x-\{x\}$, the above equals

$$\sum_{k=1}^{n}\tan^{-1}\left(\frac{1}{k}\right)-\frac{1}{k}=-\frac{\pi^{2}}{6}+\int_{0}^{1}\left(\frac{\pi\coth(\pi u)}{2u}-\frac{1}{2u^{2}}\right)du+O\left(\frac{1}{n^{2}}\right).$$ $\int_{0}^{n}\frac{x}{1+x^{2}}dx-\int_{0}^{n}\frac{\{x\}}{1+x^{2}}dx.$$Since the first integral evaluates to \sum_{k=1}^{n}\frac{1}{k}=\log n+\gamma+O\left(\frac{1}{n}\right), it follows \frac{1}{2}\log(1+x^2), we have that$$\sum_{k=1}^{n}\tan^{-1}\left(\frac{1}{k}\right)=\log n+C_{0}+O\left(\frac{1}{n}\right)$$where$$C_{0}=\gamma-\frac{\pi^{2}}{6}+\int_{0}^{1}\left(\frac{\pi\coth(\pi u)}{2u}-\frac{1}{2u^{2}}\right)du.$$n +1-\int_0^\infty \frac{\{x\}}{1+x^2}dx +O\left(\frac{1}{n}\right).$$ Edit: A negative sign was missing Be careful when using Wolfram Alpha for series which come your way. Sometimes it is wrong. Example: The series$\sum_{k=1}^\infty \frac{1}{k(k^2+u^2)}$, and the series$\sum_{k=1}^\infty \frac{1}{k^2(k^2+u^2)}$both evaluate to $$\frac{1}{2z^4}-\frac{\pi \coth(\pi z)}{2z^3}+\frac{\pi^2}{6z^2}$$ which is impossible. The latter is correct, however the former is not. This led to a small error in my previous constant. 8 added 75 characters in body; added 50 characters in body; [made Community Wiki] Asymptotics: Lets look at the quantity $$S(n)=(-1)^{n}(n+1)\binom{i}{n+1}=i\prod_{k=1}^{n}\left(1-\frac{i}{k}\right).$$ It's just your binomial coefficient above with the$(-1)^n$factored in, and an extra$n+1$so it factors nicely as a product. Claim: We have that $$S(n)=\sqrt{\frac{\sinh{\pi}}{\pi}}e^{iC_{0}+\frac{\pi}{4}}e^{i\log S(n)=\sqrt{\frac{\sinh{\pi}}{\pi}}e^{i(\frac{\pi}{4}-C_{0})}e^{-i\log n}\left(1+O\left(\frac{1}{n}\right)\right),$$ where $$C_{0}=\gamma-\frac{\pi^{2}}{6}+\int_{0}^{1}\left(\frac{\pi\coth(\pi u)}{2u}-\frac{1}{2u^{2}}\right)du\approx0.33815.$$ In particular, the angle moves around the circle like$\log n$. Application to your question: The above claim shows that $$f_{n+1} = (-1)^n n! \sqrt{\frac{\sinh{\pi}}{\pi}}\cos(\log n+C_0+\pi/4)\left(1+O\left(\frac{1}{n}\right)\right)$$ and $$g_{n+1} \sim (-1)^n n! \sqrt{\frac{\sinh{\pi}}{\pi}} \sin(\log n+C_0+\pi/4)\left(1+O\left(\frac{1}{n}\right)\right).$$ In particular, the ratio$g_n/f_n$can be made arbitrarily large or small. Proof of the claim: We first note that the size is $$\sqrt{\prod_{k=1}^{n}\left(1+\frac{1}{k^{2}}\right)}=\sqrt{\prod_{k=1}^{\infty}\left(1+\frac{1}{k^{2}}\right)}+O\left(\frac{1}{n}\right).$$ To evaluate this product, recall the Weierstrass product for the Gamma function $$\left(\Gamma(z)\right)^{-1}=ze^{\gamma z}\prod_{k=1}^{\infty}\left(1+\frac{z}{k^{2}}\right)e^{-\frac{z}{r}}.$$ From this it follows that $$\frac{1}{|\Gamma(i)|^{2}}=\frac{1}{\Gamma(i)\Gamma(-i)}=\prod_{k=1}^{\infty}\left(1+\frac{1}{k^{2}}\right).$$ Using the identity $$\Gamma(x)\Gamma(-x)=-\frac{\pi}{x\sin\left(\pi x\right)},$$ we now have that $$\frac{1}{\Gamma(i)\Gamma(-i)}=\frac{-i\sin(i\pi)}{\pi}=\frac{\sinh(\pi)}{\pi},$$ which gives rise to the$\sqrt{\frac{\sinh(\pi)}{\pi}}$term. Moving on to the evaluation of the angle, by looking at each triangle, and noting that the argument is additive when multiplied, we get that the argument equals $$\sum_{k=1}^{n}\tan^{-1}\left(\frac{1}{k}\right).$$$-\sum_{k=1}^{n}\tan^{-1}\left(\frac{1}{k}\right).$$The negative sign arises since we are working in the fourth quadrant. By looking at the Taylor series for \tan^{-1} we see that the above is \log n+O(1), however, I would like to compute this argument more precisely, and obtain the constant. Lets compare our \tan^{-1} series to the harmonic series. Notice that$$\sum_{k=1}^{n}\tan^{-1}\left(\frac{1}{k}\right)-\frac{1}{k}=-\sum_{k=1}^{n}\int_{0}^{\frac{1}{k}}\frac{x^{2}}{1+x^{2}}dx,$$Since \tan^{-1}(x)=\int_0^x \frac{1}{1+x^2}dx. The above then equals$$\sum_{k=1}^{n}\frac{1}{k}\int_{0}^{1}\frac{u^{2}}{k^{2}+u^{2}}du=\int_{0}^{1}\sum_{k=1}^{n}\frac{1}{k}\frac{u^{2}}{k^{2}+u^{2}}du.$$For |u|<1, we have that$$\sum_{k=1}^{n}\frac{1}{k}\frac{u^{2}}{k^{2}+u^{2}}=\sum_{k=1}^{\infty}\frac{1}{k}\frac{u^{2}}{k^{2}+u^{2}}+O\left(\frac{1}{n^{2}}\right),$$and we may now use the fact that this infinite series is related to the cot function function. Indeed,$$\sum_{k=1}^{\infty}\frac{1}{k}\frac{u^{2}}{k^{2}+u^{2}}=\frac{1}{2u^{2}}-\frac{\pi\coth(\pi u)}{2u}+\frac{\pi^{2}}{6},$$and so$$\sum_{k=1}^{n}\tan^{-1}\left(\frac{1}{k}\right)-\frac{1}{k}=-\frac{\pi^{2}}{6}+\int_{0}^{1}\left(\frac{\pi\coth(\pi u)}{2u}-\frac{1}{2u^{2}}\right)du+O\left(\frac{1}{n^{2}}\right).$$Since \sum_{k=1}^{n}\frac{1}{k}=\log n+\gamma+O\left(\frac{1}{n}\right), it follows that$$\sum_{k=1}^{n}\tan^{-1}\left(\frac{1}{k}\right)=\log n+C_{0}+O\left(\frac{1}{n}\right)$$where$$C_{0}=\gamma-\frac{\pi^{2}}{6}+\int_{0}^{1}\left(\frac{\pi\coth(\pi u)}{2u}-\frac{1}{2u^{2}}\right)du.$$Acknowledgements: I would like to thank Noam Elkies for pointing out that$$\prod_{k=1}^\infty \sqrt{1+\frac{1}{k^2}}=\frac{1}{|\Gamma(i)|}=\sqrt{\frac{\sinh(\pi)}{\pi}} in the comments.

Edit:

A negative sign was missing.

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