1

Let's consider sum:

$$L (x_n) = \sum_{n=1}^\infty \sin(n) \cdot (x_n - x_{n+1})$$

For $x_n$ such that $\sum_{n=1}^\infty|x_n| \le 1$

I want to check whether this functional is well defined i.e. to check if $\forall x_n$ this sum exists.

As already was shown here: Maximum of $\sum_{n=1}^\infty \sin(n) \cdot (x_n - x_{n+1})$ we know that $L(x_n)$ is bounded, but I'm not sure how to show that this sum exists for any $x_n$ i.e. to show that for any $x_n$ matching the condition $\sum_{n=1}^\infty|x_n|\le1$ $,L(x_n)$ converges.

I was trying to prove it (combining facts that we have $(x_n - x_{n+1})$ differences which strongly suggests to use Cauchy's theorem and condition on $x_n$ : $\sum_{n=1}^\infty |x_n| \le 1$). But I end up with nothing. Could you please give me a hand with telling if this expression converges for any $x_n$ ?

Lucian
  • 1,807

2 Answers2

2

Yes, this series converges absolutely. This is because $|\sin(n) (x_{n + 1} - x_n)| \leq |x_{n + 1} - x_n| \leq |x_{n + 1}| + |x_n|$, and

$\sum\limits_{n = 1}^\infty |x_{n + 1}| + |x_n|$

converges.

Mark Saving
  • 33,541
0

For any $\epsilon>0$, just choose $N_0$ large enough such that $\sum\limits_{n=N_0}^{N_1}|x_n| <\frac1{2}\epsilon$ for all $N_1 \gt N_0$

and you get $\sum\limits_{n=N_0}^{N_1}|\sin(n)(x_n-x_{n+1})| \le \sum\limits_{n=N_0}^{N_1}|\sin(n)x_n| +\sum\limits_{n=N_0}^{N_1}|\sin(n) x_{n+1}| < \epsilon$

so $\left|\sum\limits_{n=N_0}^{N_1}\sin(n)(x_n-x_{n+1})\right| < \epsilon$

so the partial sums are a Cauchy sequence

and thus the overall series converges.

Henry
  • 169,616