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Let $\{P_n\}$ be a sequence of points determined as in the figure. Find $\lim_{n \to \infty} \angle P_nAP_{n+1}. $

$|P_nP_{n+1}| = 2^{n-1}$, however $|AP_n|$ I cannot find an expression for because you have to define it's the hypotenuse of the previous triangle, and to define that you need to find the hypotenuse of the triangle before, etc.

Even then I have no clue how to do this. Perhaps if you define $|AP_n|$ you can use a trig function to find a limit, but as I said I do not know how to define $|AP_n|$. How to do this problem?

y

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    Hint: Set $h_n=A_1P_n$. You have, using Pythagoras $h_{n+1}^2=h_n^2+(2^{n-2})^2$. Set $H_n=h_n^2$. $(H_n)$ is a arithmetical sequence (or progression). Thus you can have an explicit expression for $H_n$... then for $h_n$... – 2017-01-21
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    @JeanMarie It seems that $H_{n+1} = 4H_n - 2$ – 2017-01-21

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The hypothenuses grow with the recurrence

$$f (n+1)=\sqrt {f (n)^2+2^{n-1}} $$

beginning with $f (n)=1\,$. This recurrence, whose first values are $\sqrt {2} $, $\sqrt {6} $, $\sqrt {22} $, $\sqrt {86} $... has solution

$$\sqrt {\frac { 2^{2n}+2 }{3} }$$

So the ratio $R $ between the external leg and the hypothenuse is

$$R=\frac {2^{n-1}}{ \sqrt {\frac { 2^{2n}+2 }{3}} }$$

Taking the limit for $n \rightarrow \infty $, we get that $R $ tends to $\frac { \sqrt {3}}{2}\, $.

Since the angle cited in the OP corresponds to $\arcsin (R) \,\,\,$ and must be acute, we conclude that the limit of the angle is $\frac {\pi}{3}\, $.

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    How did you find the solution to the recurrence relation? – 2017-01-21
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    Noting that the sequence, as defined, can be written as $\sqrt {2} $, $\sqrt {2+4} $, $\sqrt {2+4+16} $, $\sqrt {2+4+16} $... So the $n^{th} $ term is equivalent to the partial sum $\sqrt {2+\sum 4^{n-1}} $. Since it is known that $$\sum k^{n-1}=\frac {k}{k-1}(k^{n-1}-1) $$ we get $$\sqrt {2+ \frac {4}{3}(4^{n-1}-1) } $$ and then $$\sqrt {2+ \frac {1}{3}(2^{2n}-4) } $$ $$=\sqrt { \frac {1}{3}(2^{2n}+2) } $$ – 2017-01-21
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    I am not familiar with the sum you use, what is $\sum 4^{n-1}$? From where to where does it sum? I know the formula $\sum_{i=0}^n k^n = \dfrac{k^{n+1}-1}{k-1}$ – 2017-01-21
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    Sorry for not having better specified. The sequence is $\sqrt {2+\sum_{i=2}^n 4^{n-1} } $. To calculate the sum under the radical, we can start from the formula that you cited, noting that it implies $$\sum_{i=1}^{n} k^{n-1}=\frac {k^n-1}{k-1} $$ Since we are interested in this sum starting from $i=2$, we have to subtract the term corresponding to $i=1$, which is $k^0=1$. So we have $$\sum_{i=2}^{n} k^{n-1}=\frac {k^n-1}{k-1} -1\\ =\frac {k^n-k}{k-1} $$ Setting $k=4$ we get $$\sum_{i=2}^{n} 4^{n-1}= \frac {4^n-4}{3} =\frac {4}{3} (4^{n-1}-1 ) $$ – 2017-01-21
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    Thank you very much, excellent. – 2017-01-22