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So in analysis 1 we re doing limits of sums and general terms and so on and this is one of the exercises we have at homework. And just like most exercises that involve fractional parts, i dont even know where to start it from

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

Any suggestions?

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  • $\begingroup$ For positive $x$ we have $$\{x\}=\frac{1}{2}-\frac{\sum _{k=1}^{\infty } \frac{\sin (2 \pi k x)}{k}}{\pi }$$ $\endgroup$ – Raffaele Oct 24 '20 at 12:21
  • $\begingroup$ @Raffaele Can you please give us a source? $\endgroup$ – PNDas Oct 24 '20 at 12:25
  • $\begingroup$ @PNDas functions.wolfram.com/IntegerFunctions/FractionalPart/06/01 $\endgroup$ – Raffaele Oct 24 '20 at 12:28
  • $\begingroup$ Try to express $(1+\sqrt2)^{2n-1}$ differently. $\endgroup$ – Yves Daoust Oct 24 '20 at 13:54
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Let $a_n$ be defined by

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

Then it is straightforward to verify that

$$ a_0 = a_1 = 2, \qquad a_{n+2} = 2a_{n+1} + a_n. $$

In particular, $a_n$ is integer for all $n \geq 0$. So, if $n \geq 1$, then

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

This and $0 < \sqrt{2}-1 < 1$ together then show that

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

Therefore

$$ \sum_{n=1}^{\infty} (\sqrt{2}-1)^{2n-1} = \frac{\sqrt{2}-1}{1 - (\sqrt{2}-1)^2} = \frac{1}{2}. $$

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This is actually a simple geometric progression, in disguise.

The first term in the sequence is $1+\sqrt2$ and its fractional part is $\sqrt2-1$. But $$(\sqrt2+1)(\sqrt2-1)=1$$

Hence for any integer $k$, $$(\sqrt2+1)^k(\sqrt2-1)^k=1$$ and the fractional part of each $(\sqrt2+1)^k$ is $(\sqrt2-1)^k$ when $k$ is odd.

Now $(1\pm\sqrt2)^2=3\pm2\sqrt2$, so the terms in the series

$$\{(1+\sqrt2)^{2n-1}\}$$ are simply

$$(\sqrt2-1), (\sqrt2-1)(3-2\sqrt2), (\sqrt2-1)(3-2\sqrt2)^2,\dots$$

That is, we have a geometric progression with first term $\sqrt2-1$ and ratio $3-2\sqrt2$, so its sum is

$$\begin{align}\\ & \frac{\sqrt2-1}{1-3+2\sqrt2}\\ & =\frac{\sqrt2-1}{2(\sqrt2-1)}\\ & =\frac12\\ \end{align}$$

FWIW, these terms are related to the solutions of the (so-called) negative Pell's equation $$x^2 - 2y^2 = -1$$ which can be factored as $$(x+\sqrt2y)(x-\sqrt2y)=-1$$ The first few (x,y) tuples are (1,1), (7,5), (41, 29), (239, 169).

Since $$(1+\sqrt2)(1-\sqrt2)=-1$$ $$(1+\sqrt2)^k(1-\sqrt2)^k=-1^k$$ and we need $k$ to be odd to get the negative solutions required for the above problem.

The solution tuples for odd & even $k$ are (1,1),(3,2),(7,5),(17,12),(41,29), etc. There's a very simple pattern here: $$x_{k+1}=x_k+2y_k$$ $$y_k=x_k+y_k$$ Also, $$x_{k+1}=2x_k+x_{k-1}$$ $$y_{k+1}=2y_k+y_{k-1}$$

When $k$ is odd, we get $$2y^2-x^2=1$$ as desired. Eg, for $k=3$, we have $$2\cdot5^2-7^2=1$$ that is $$(\sqrt{50}+\sqrt{49})(\sqrt{50}-\sqrt{49})=1$$

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    $\begingroup$ Your answer is correct, but I fail to see why the computation $(\sqrt{2}+1)^k(\sqrt{2}-1)^k=1$ leads to the claim $\{(\sqrt{2}+1)^k\}=(\sqrt{2}-1)^k$. Indeed, this is true only when $k$ is an odd integer, and in general $$\{(\sqrt{2}+1)^k\}=\begin{cases}(\sqrt{2}-1)^k,&\text{if $k$ is odd}, \\ 1-(\sqrt{2}-1)^k,&\text{if $k$ is even}\end{cases}.$$ $\endgroup$ – Sangchul Lee Oct 24 '20 at 13:19
  • $\begingroup$ @SangchulLee Thanks. Yes, I was a bit sloppy about that. I was in a rush to submit my answer (& I'm on my phone). I'll try to tighten things up shortly. $\endgroup$ – PM 2Ring Oct 24 '20 at 13:33

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