Sum of fractional parts 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?
 A: 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
$$ \sum_{n=1}^{\infty} (\sqrt{2}-1)^{2n-1} = \frac{\sqrt{2}-1}{1 - (\sqrt{2}-1)^2} = \frac{1}{2}. $$
A: 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
$$(\sqrt{50}+\sqrt{49})(\sqrt{50}-\sqrt{49})=1$$
