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Consider the sequence of squares and angles as in this figure:

enter image description here

Since $\tan \alpha_n=\frac{1}{n}$, we can show that $\alpha_1+\alpha_2+\alpha_3=\frac{\pi}{2}$ (see :Determine the angle of 3 drawn lines from each corner of 3 congruent squares)

For $n>3$ the sum of the angles $$ \beta_n= \sum_{i=1}^n \alpha_i $$ becomes more difficult to find using the tigonometric formulas for the tangent of sum angles. So, the first question is if there is some other method to find this sum.

The second question is if the series $$ \sum_{i=1}^\infty \alpha_i $$ converge or not and, if it converges, what is the sum.

I've tempted a numerical experiment that gives: $\beta_{1000}=\pi \cdot2.294981074\ldots$ and $\beta_{2000}=\pi \cdot2,515537077\ldots$ But, obviously, this is not conclusive.

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    $\begingroup$ for large $n$, $\tan\alpha_n = \frac1n \implies \alpha_n \approx \frac1n$, the partial sums $\sum_{k=1}^N \alpha_k$ diverges like $\log N$. $\endgroup$ – achille hui Dec 14 '15 at 22:19
  • $\begingroup$ It's fairly easy to show this infinite series diverges. See my answer below. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 15 '15 at 2:03
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Notice for real number $x$, we have $$\tan^{-1}(x) = \Im\log(1+ix) = \frac{1}{2i}\log\left(\frac{1+ix}{1-ix}\right)$$

This gives us

$$\begin{align} \alpha_n = \tan^{-1}\frac1n &= \frac{1}{2i}\log\left(\frac{n+i}{n-i}\right) = \frac{1}{2i}\log\left(\frac{\Gamma(n+1+i)/\Gamma(n+i)}{\Gamma(n+1-i)/\Gamma(n-i)}\right)\\ &= \frac{1}{2i}\left[\log\left(\frac{\Gamma(n+1+i)}{ \Gamma(n+1-i)}\right)-\log\left(\frac{\Gamma(n+i)}{\Gamma(n-i)}\right)\right]\\ &= \Im\log\Gamma(n+1+i) - \Im\log\Gamma(n+i) \end{align} $$ where $\Gamma(x)$ is the Gamma function.

A corollary of this is the partial sums has a closed form expression: $$\sum_{k=1}^N \alpha_k = \Im\log\Gamma(N+1+i) - \Im\log\Gamma(1+i)\tag{*1}$$

For $|\arg z | \le \pi - \delta$, $\log\Gamma(z)$ has following asymptotic expansion for large $|z|$:

$$\log\Gamma(z) \asymp (z- \frac12)\log z - z + \frac12\log(2\pi) + \sum_{r=1}^\infty \frac{B_{2r}}{2r(2r-1)z^{2r-1}}\tag{*2}$$ where $B_r$ are the Bernoulli numbers.

Replace $z$ by $N + 1 \pm i$ and apply $(*2)$ to $(*1)$, one find the partial sums diverges like $\log N$ plus a constant offset $-\Im\log\Gamma(1+i) \approx 0.30164032046$:

$$\sum_{k=1}^N \alpha_k = \log(N) -\Im\log\Gamma(1+i) + \frac{1}{2N}+\frac{1}{12N^2}-\frac{1}{6N^3}+O(N^{-4}) $$

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Since $\beta_n = \beta_{n-1}+\alpha_n$, we have $\tan \beta_n = \tan(\beta_{n-1}+\alpha_n) = \dfrac{\tan \beta_{n-1} + \tan \alpha_n}{1-\tan\beta_{n-1}\tan\alpha_n} = \dfrac{\tan \beta_{n-1} + \tfrac{1}{n}}{1-\tfrac{1}{n}\tan\beta_{n-1}}$.

This gives us a recursive relation for $\tan \beta_n$. I believe this is the best we can do.

As already mentioned in the comments, for large $n$, $\alpha_n = \arctan\dfrac{1}{n} \sim \dfrac{1}{n}$, so the partial sums will behave like $\beta_N \sim \log N$ for large $N$. If you try you're numerical experiment for larger $N$, you will get $\beta_{10^4} \approx 3.02777\pi$, $\beta_{10^5} \approx 3.76069\pi$, $\beta_{10^6} \approx 4.49363\pi$, $\beta_{10^7} \approx 5.22656\pi$, and $\beta_{10^8} \approx 5.95950\pi$. As you can see, this is growing logarithimically as predicted.

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$$ \tan \alpha_n = \frac 1 n. $$ $$ \frac x {\tan x} \to 1 \text{ as }x\to0. $$ Hence $$ \frac{\alpha_n}{1/n} = \frac{\alpha_n}{\tan\alpha_n} \to 1 \text{ as }n\to\infty. $$ If $a_n,b_n>0$ for all $n$ and $\lim\limits_{n\to\infty}\dfrac{a_n}{b_n}$ is a strictly positive number, then the two series $\sum\limits_{n=1}^\infty a_n$ and $\sum\limits_{n=1}^\infty b_n$ either both converge or both diverge. We know that $\sum\limits_{n=1}^\infty \frac 1 n$ diverges. Therefore the other series also diverges.

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We can use the Integral Test for Convergence:

Assume the series converges to a real number, then:

$$\int_1^\infty \tan^{-1} \left(\frac{1}{x}\right) \, dx \leq \sum_{n=1}^\infty \tan^{-1} \left(\frac{1}{n}\right) \leq \tan^{-1}(1) + \int_1^\infty \tan^{-1} \left(\frac{1}{x}\right) \, dx$$

where

$\int\tan^{-1}\big(\frac{1}{x}\big)dx = \ln(x^2+1)+x\tan^{-1}\big(\frac{1}{x}\big)+c$

but

$\lim_{x\to\infty}\ln(x^2+1)+x\tan^{-1}\big(\frac{1}{x}\big) \notin \mathbb{R}$

so the integral does not converge, and thus, the series also does not converge.

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  • $\begingroup$ I think you may have forgotten a factor of $\frac{1}{2}$ in front of $\ln(x^2+1)$. $\endgroup$ – Strants Dec 15 '15 at 3:58
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    $\begingroup$ Notice that we can actually use this technique to bound the partial sums $\beta_n$. In particular, $$\int_1^{N+1} \tan^{-1}\frac{1}{x}\;dx \le \beta_n \le \tan^{-1}(1) + \int_1^N \tan^{-1}\frac{1}{x}\;dx$$ $\endgroup$ – Strants Dec 15 '15 at 4:00

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