Pointwise convergence for all real x

[lys04]

Homework Statement

pointwise convergence of a series

Relevant Equations

$$ \sum_{n=1}^\infty sin(\frac{n\pi}{2})sin(nx) $$

How do I know whether or not the series
$$ \sum_{n=1}^\infty sin(\frac{n\pi}{2})sin(nx)$$

converges pointwise for all real x or not?

By the way am I right in thinking that converging pointwise for all real x means whatever x i plug into the series it converges to some finite value?

I was thinking if i plug in x=pi/2 then I'd get

$$ \sum_{n=1}^\infty sin^2(\frac{n\pi}{2}) $$

Which diverges, does that prove that the series doesn't converge pointwise for all x?

 

[Orodruin]

How do I know whether or not the series
$$ \sum_{n=1}^\infty sin(\frac{n\pi}{2})sin(nx)$$

converges pointwise for all real x or not?

You either show that it does or find a way to show that it does not.

By the way am I right in thinking that converging pointwise for all real x means whatever x i plug into the series it converges to some finite value?

Yes.

I was thinking if i plug in x=pi/2 then I'd get

$$ \sum_{n=1}^\infty sin^2(\frac{n\pi}{2}) $$

Which diverges, does that prove that the series doesn't converge pointwise for all x?

Yes.

 

[PeroK]

Homework Statement: pointwise convergence of a series
Relevant Equations: $$ \sum_{n=1}^\infty sin(\frac{n\pi}{2})sin(nx) $$

How do I know whether or not the series
$$ \sum_{n=1}^\infty sin(\frac{n\pi}{2})sin(nx)$$

converges pointwise for all real x or not?

Note that $\sin(\frac{n\pi}{2}) = 0$ if $n$ is even and alternates between $\pm1$ if $n$ is odd. Using the alternating series test, $\sum_{n=1}^\infty sin(\frac{n\pi}{2})a_n$ converges iff $a_n \to 0$ for odd $n$.

In this case, the series converges iff $\sin((2n+1)x) \to 0$ as $n \to \infty$. That's only the case when $x = k\pi$ for some integer $k$. Although, it's not that easy from first principles to prove that it does not converge for any other $x$. You found a good example,

 

[fresh_42]

In this case, the series converges iff $\sin((2n+1)x) \to 0$ as $n \to \infty$. That's only the case when $x = k\pi$ for some integer $k$. Although, it's not that easy from first principles to prove that it does not converge for any other $x$. You found a good example,

The series expansion of the sine function can help. Not really first principles but at least in the realm of the question.

Another idea is to substitute $y=x-(\pi/2)$ which turns the series members into $\sin^2(n \pi /2)\cos(ny)$ which reduces the question to: Under what circumstances does $\sum \sin^2(n \pi /2)\cos(ny)$ converge, if $\sum \sin^2(n \pi /2)$ diverges?

One should also keep the Weierstraß- or half-angle substitution in mind whenever it comes to trig functions and integrals or sums.

 

[Office_Shredder]

Note that $\sin(\frac{n\pi}{2}) = 0$ if $n$ is even and alternates between $\pm1$ if $n$ is odd. Using the alternating series test, $\sum_{n=1}^\infty sin(\frac{n\pi}{2})a_n$ converges iff $a_n \to 0$ for odd $n$.

One small point, we don't actually know the $a_n$ are positive in this question so the alternating test doesn't work.

 

[PeroK]

One small point, we don't actually know the $a_n$ are positive in this question so the alternating test doesn't work.

Yes, well spotted.