# Simple Convergence

bomba923
Just out of curiosity, does
$$\sum\limits_{n = 1}^\infty {\sin \left( {x^n } \right)}$$
converge $\forall x \in \left( -1 , 1 \right)$ ?

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Homework Helper
Gold Member
I would be tempted to say yes, because of the inequality |sin(x)|<|x|, which implies |sin(x^n)|<|x^n|, which implies

$$\sum^m|\sin(x^n)|\leq \sum^m|x^n|$$

But for -1<x<1, the sum on the right is geometric, which converges. So the sine series converges absolutely.

murshid_islam
can you please tell me how i can prove that |sin(x)| < |x|?

Homework Helper
Strictly speaking, it's not true- sin(x)= x when x= 0!

At x=0 sin(0)= 0. Let f(x)= x- sin(x). Then f'(x)= 1- cos(x) which is always greater than or equal to 0. That is, x- sin(x) is an increasing function. For any x> 0 x> sin(x). For x< 0 use the fact that both x and sin(x) are odd functions: if x is negative, then -x is positive and so -x> sin(-x)= -sin(x). But x is negative and (for $0> x> -\pi$) so is sin(x). |sin(x)|= -x> |sin(x)|= -sin(x). For |x|> 1, it is obvious that |x|> |sin(x)|.

bomba923
I would be tempted to say yes, because of the inequality |sin(x)|<|x|, which implies |sin(x^n)|<|x^n|, which implies

$$\sum^m|\sin(x^n)|\leq \sum^m|x^n|$$

But for -1<x<1, the sum on the right is geometric, which converges. So the sine series converges absolutely.
I see, thank you

A similar question (but like the previous, related to a larger problem):
$$\forall n \in \mathbb{N},{\text{ does }}\exists x > 1:\frac{1} {n}\sum\limits_{k = 1}^n {\sin \left( {x^k } \right)} \leqslant \sin 1{\text{ ?}}$$

bomba923
In particular, I'm looking at the function
$$f_n \left( x \right) = \frac{1}{n}\sum\limits_{k = 1}^n {\sin \left( {x^k } \right)}$$

where
$$\forall n \geqslant 1,\; - 1 \leqslant f_n \left( x \right) \leqslant 1$$
and
$$\forall x \in \left[ 0,1 \right], \; \frac{{df_n }}{{dx}} > 0$$

From inspection at n=2,3,..., up to n=25 (my graphing program's constraints!),
the first critical point (i.e., dfn(x)/dx=0) that fn(x) approaches (for positive x) is always a maximum at some x>1

Furthermore, if we let xn be the x-value of this maximum for fn(x)
(i.e., the smallest x>0 such that dfn(x)/dx=0), then we notice
$$x_2 > \cdots > x_n > 1$$
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*But, what is
$$\lim \limits_{n \to \infty } x_n \; ?$$

Clearly, $\lim \limits_{n \to \infty } x_n \ne 1$, since
$$f'_n \left( 1 \right) = \left( {\cos 1 + 2\cos 1 + 3\cos 1 + \cdots } \right)/n > 0$$

Also, since
$$\frac{{df_n }}{{dx}} = \frac{1}{n}\sum\limits_{k = 1}^n {kx^{k - 1} \cos \left( {x^k } \right)}$$

my question is simply
$${\text{What is the smallest }}x > 0\;{\text{such that }}\frac{d} {{dx}}\sum\limits_{k = 1}^{\infty} {\sin \left( {x^k } \right)} = 0\;?$$

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bomba923
No replies?
A similar question (but like the previous, related to a larger problem):
$$\forall n \in \mathbb{N},{\text{ does }}\exists x > 1:\frac{1} {n}\sum\limits_{k = 1}^n {\sin \left( {x^k } \right)} \leqslant \sin 1{\text{ ?}}$$
$$\text{What is } \lim \limits_{n \to \infty } x_n ?$$

$${\text{How do we know if }}\forall n \in \mathbb{N},\;\exists x_n > 1: f_n {\kern 1pt} ' \left( {x_n } \right) = 0\;?$$

which means I must
$${\text{Prove/disprove that }}\forall n \in \mathbb{N},\;\exists x_n > 1:\frac{d}{{dx}}\sum\limits_{k = 1}^n {\sin \left( {x_n^k } \right)} = 0$$

or, equivalently (due to Mean & Intermediate Value Theorems),
$${\text{Prove/disprove that }}\forall n \in \mathbb{N},\;\exists x > 1: \frac{1}{n} \sum\limits_{k = 1}^n {\sin \left( {x^k } \right)} \leqslant \sin 1$$

(I tried induction, but showing n$\to$n+1 wasn't quite as easy as I hoped...)

*So, does anyone have any ideas how I may prove (or disprove ) that
$$\forall n \in \mathbb{N},\;\exists x > 1: \frac{1}{n} \sum\limits_{k = 1}^n {\sin \left( x^k \right)} \leqslant \sin 1$$
?

bomba923
Same as before,
$$\begin{gathered} f_n \left( x \right) = \frac{1} {n}\sum\limits_{k = 1}^n {\sin \left( {x^k } \right)} \Rightarrow f'_n \left( x \right) = \frac{1} {n}\sum\limits_{k = 1}^n {kx^{k - 1} \cos \left( {x^k } \right)} \hfill \\ x_n = \min \left\{ {x > 0:f'_n \left( x \right) = 0} \right\} = \min \left\{ {x > 0:\frac{d} {{dx}}\sum\limits_{k = 1}^n {\sin \left( {x^k } \right)} = 0} \right\} \hfill \\ \end{gathered}$$

From inspection, it appears that
$$\frac{d}{{dx}}\sum\limits_{k = 1}^{n + 1} {\sin \left( {x_n^k } \right)} < 0$$

though...
$$\begin{gathered} \cos x_n + 2x_n \cos x_n^2 + 3x_n^2 \cos x_n^3 + \cdots + nx_n^{n - 1} \cos x_n^n = 0 \Rightarrow \hfill \\ \cos x_n + 2x_n \cos x_n^2 + 3x_n^2 \cos x_n^3 + \cdots + \left( {n + 1} \right)x_n^n \cos x_n^{n + 1} < 0 \hfill \\ \end{gathered}$$

~or equivalently,
$$\begin{gathered} \cos x_n + 2x_n \cos x_n^2 + 3x_n^2 \cos x_n^3 + \cdots + nx_n^{n - 1} \cos x_n^n = 0 \Rightarrow \hfill \\ \cos x_n + 2x_n \cos x_n^2 + 3x_n^2 \cos x_n^3 + \cdots + nx_n^{n - 1} \cos x_n^n + \left( {n + 1} \right)x_n^n \cos x_n^{n + 1} < 0 \hfill \\ \end{gathered}$$

implies (as π/2=x1>x2>...)
$$\left( {n + 1} \right)x_n^n \cos x_n^{n + 1} < 0 \Rightarrow \cos x_n^{n + 1} < 0 \Rightarrow \sqrt[{n + 1}]{{\pi /2}} < x_n < \sqrt[{n + 1}]{{3\pi /2}}$$

which, by the Squeeze Theorem, implies
$$\lim \limits_{n \to \infty } \sqrt[{n + 1}]{{\pi /2}} < \lim \limits_{n \to \infty } x_n < \lim \limits_{n \to \infty } \sqrt[{n + 1}]{{3\pi /2}} \Rightarrow \mathop {\lim }\limits_{n \to \infty } x_n = 1$$

*Which is strange, considering that
$$\begin{gathered} \forall n \in \mathbb{N},\;f'_n \left( x \right) > 0\;{\text{for }}0 \leqslant x \leqslant 1,\;{\text{and}} \hfill \\ \mathop {\lim }\limits_{n \to \infty } f'_n \left( 1 \right) = \frac{{\cos 1 + 2\cos 1 + 3\cos 1 + \cdots + n\cos 1}} {n} = \mathop {\lim }\limits_{n \to \infty } \frac{{n + 1}} {2}\cos 1,\;{\text{which diverges}} \hfill \\ \end{gathered}$$

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