limit

[kinu]

\(\displaystyle \lim_{n\rightarrow \infty}\left\{\left(\sqrt{2}+1\right)^n\right\}.(-1)^{\left[\left(\sqrt{2}+1\right)^n\right]}\)

Where \(\left\{x\right\}=\) fractional part function and \(\left[x\right]=\) Integer part function

and \(n\in \mathbb{N}\)

[Rui Carpentier]

Hint:
Show first that \((1+\sqrt{2})^n+(1-\sqrt{2})^n\) is an even number.
Then conclude that \([(\sqrt{2}+1)^n]\) is odd if \(n\) is even and even if \(n\) is odd (note that \(-1<1-\sqrt{2}<0\)).
After that the exercise is easy.

[kinu]

Thanks Sir.

my process::

I have used two cases

(i) When \(n=2m\) and \(m\in \mathbb{N}\) means \(n\) is even

(ii) When \(n=2m+1\) and \(m\in \mathbb{N}\) means \(n\) is odd


Now for (i)

\(\lim_{m\rightarrow \infty} \left\{\left(\sqrt{2}+1\right)^{2m}\right\}\times (-1)^{\left[(\sqrt{2}+1)^{2m} \right]}\)

Now \(\left(\sqrt{2}+1\right)^{2m} =\left [\left (\sqrt{2}+1\right)^{2m} \right\]+\left\{\left(\sqrt{2}+1\right)^{2m}\right\}\)

Now Let \(\left\{\left(\sqrt{2}+1\right)^{2m}\right\} = f\Leftrightarrow 0 \leq f<1\)

Now Let \(\left\{\left(\sqrt{2}-1\right)^{2m}\right\} = f^{'}\Leftrightarrow 0<f<1\)

So \(\left\{\left(\sqrt{2}+1\right)^{2m}\right\} = \left(\sqrt{2}-1\right)^{2m}\)

So \(\left(\sqrt{2}+1\right)^{2m}+\left(\sqrt{2}-1\right)^{2m} = 2N\;\;, n\in \mathbb{N}\)

So \(\left [\left(\sqrt{2}+1\right)^{2m} \right]+f+f^{'} = 2N\)

So \(f+f^{'}\in \mathbb{N}\Leftrightarrow f+f^{'} = 1\)

So \(\lim_{m\rightarrow \infty} \left(\sqrt{2}+1\right)^{2m}+\lim_{m\rightarrow \infty} \left(\sqrt{2}+1\right)^{2m} = 1\)

So \(\lim_{m\rightarrow \infty} \left(\sqrt{2}+1\right)^{2m}+0\) = 1

and \(\left [\left(\sqrt{2}+1\right)^{2m} \right] = 2N-1\)

So \(\displaystyle \lim_{n\rightarrow \infty}\left\{\left(\sqrt{2}+1\right)^n\right\}.(-1)^{\left[\left(\sqrt{2}+1\right)^n\right]} = 1\times (-1)^{2N-1} = -1\)

But How can I calculate for (ii) case

Help Required..

[Rui Carpentier]

All you have to do is to show that:

\((-1)^{[(\sqrt{2}+1)^n]}=\left\{\begin{array}{cc}+1 & \mbox{if } n \mbox{ is odd}\\ -1 & \mbox{if } n \mbox{ is even}\end{array} \right.\)

and

\(\{(\sqrt{2}+1)^n\}=\left\{\begin{array}{cc}(\sqrt{2}-1)^n & \mbox{if } n \mbox{ is odd}\\ 1-(\sqrt{2}-1)^n & \mbox{if } n \mbox{ is even}\end{array} \right.\)

Both came from the fact that

\((\sqrt{2}+1)^n=\left\{\begin{array}{cc}2k+(\sqrt{2}-1)^n & \mbox{if } n \mbox{ is odd}\\ 2k-(\sqrt{2}-1)^n & \mbox{if } n \mbox{ is even}\end{array} \right.\)

which came from the fact that \((1+\sqrt{2})^n+(1-\sqrt{2})^n=2k\) (an even number) and \(0<(\sqrt{2}-1)^n<1\).

With this you shall complete the exercise.