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I am trying the following exercise,

Convergence of the series $\sum_{n=0}^{+\infty} \sin((1+\sqrt{2})^n\pi)$

I tried like the method for $\sum_{n=0}^{+\infty} \sin((2+\sqrt{3})^n\pi)$,

with $u_n=\sin((1-\sqrt{2})^n\pi)$ unfortunately $(1-\sqrt{2})<0$ so I cannot use theorem for positivness.

There is an another trick for this one ?

Thank you in advance

Free X
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1 Answers1

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Let $b_n=(1+\sqrt{2})^n +(1-\sqrt{2})^n$. Then $b_n$ is an integer. (As pointed out by Aaron, that can be seen by expanding, using the Binomial Theorem, and noting the cancellations. It can also be seen by noting that the expression for $b_n$ is invariant under the mapping that takes $\sqrt{2}$ to $-\sqrt{2}$.)

So $\pi(1+\sqrt{2})^n$ differs from $\pi b_n$ ($\pi$ times an integer) by $\pi(1-\sqrt{2})^n$. Note that if $n$ is large, then the absolute value of $\pi(1-\sqrt{2})^n$ is very close to $0$.

The $n$-th term of our series has absolute value $|\sin(\pi(1-\sqrt{2})^n)|$, which is less than $\pi(\sqrt{2}-1)^n$.

But $\sum \pi(\sqrt{2}-1)^n$ converges, since it is a geometric series with common ratio $\lt 1$. Thus by Comparison the series of the problem converges absolutely, and hence converges.

Remark: Whenever $1+\sqrt{2}$ has a problem, its buddy $1-\sqrt{2}$ is ready to help.

André Nicolas
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  • Note that the assertion in the first sentence can be seen by expanding both powers using the binomial theorem and noting that the terms with odd powers of $\sqrt{2}$ cancel out. – Aaron Apr 20 '14 at 20:10
  • Of course, I forgot to factorize my series to get a geometric series. Thanks – Free X Apr 20 '14 at 20:13
  • (What did you mean by the expression is invariant under the mapping ?) –  Apr 20 '14 at 20:17
  • Another way to prove that $u_n := (1+\sqrt{2})^n+(1-\sqrt{2})^n$ is an integer comes from the fact that $(1+\sqrt{2})$ and $(1-\sqrt{2})$ are conjugated quadratics : they are both roots of the polynomial $x^2-2x-1$. Hence, it can be shown that $u_{n+2} = 2u_{n+1}+u_n$, and $u_0 = u_1 = 2$, from where a trivial reccurence shows that $u_n$ is always an integer. – D. Thomine Apr 20 '14 at 20:19
  • Elaborating on the invariance, the expression clearly lies in the field $\mathbb Q[\sqrt 2]$, and being invariant under the conjugation means (by basic galois theory) that the expression is rational. However, it is also an algebraic integer (being a sum of powers of algebraic integers), and every rational algebraic integer is in fact an integer. – Aaron Apr 20 '14 at 20:20
  • If everywhere in $(1+\sqrt{2})^n +(1-\sqrt{2})^n$ we replace $\sqrt{2}$ by $-\sqrt{2}$, the two terms get interchanged, but the sum does not change. – André Nicolas Apr 20 '14 at 20:20
  • @D.Thomine And you can do like this for several series too..like $\frac{1+\sqrt{5}}{2}$, André Nicolas thanks! –  Apr 20 '14 at 20:21
  • It seems important to note that $\left(1+\sqrt2\right)^n+\left(1-\sqrt2\right)^n$ is an even integer since that makes it possible to say $\sin\left(\left(1+\sqrt2\right)^n\pi\right)=(-1)^{n+1}\sin\left(\left(\sqrt2-1\right)^n\pi\right)$ – robjohn Jun 29 '19 at 19:24