Does the series Ʃ sinx / x converge or diverge?

[fakecop]

Homework Statement

Determine whether the series Ʃ(1 to infinity) sinx / x converges or diverges.

Homework Equations

This question appears in the integral test section, but as far as i know the integral test can only be used for decreasing functions, right?

The Attempt at a Solution

Using the ratio test, limx->infinity sin(x+1)/(x+1)*(sinx/x)=limx->infinity x*sin(x+1)/((x+1)(sinx))
This is where i got stuck-this limit oscillates between positive infinity and negative infinity.
Using the root test, i need to find the limit of (sinx)^(1/x) as x approaches infinity, which also gets me nowhere.

we have not done taylor series yet so I'm sure there is a relatively simple approach to this question...please help?

 

[Dick]

Homework Statement

Determine whether the series Ʃ(1 to infinity) sinx / x converges or diverges.

Homework Equations

This question appears in the integral test section, but as far as i know the integral test can only be used for decreasing functions, right?

The Attempt at a Solution

Using the ratio test, limx->infinity sin(x+1)/(x+1)*(sinx/x)=limx->infinity x*sin(x+1)/((x+1)(sinx))
This is where i got stuck-this limit oscillates between positive infinity and negative infinity.
Using the root test, i need to find the limit of (sinx)^(1/x) as x approaches infinity, which also gets me nowhere.

we have not done taylor series yet so I'm sure there is a relatively simple approach to this question...please help?

The answer to this is actually not very simple. You need something like http://en.wikipedia.org/wiki/Dirichlet's_test Have you covered anything like that? It's a generalization of the alternating series test. None of the elementary tests will work. As you've figured out.

 

[fakecop]

Thank you for your reply. The text we are using only covers the elementary tests, and the textbook solved this question via the integral test, which leads me to believe that the book has made a mistake as this is not a monotonic decreasing function.

 

[Dick]

Thank you for your reply. The text we are using only covers the elementary tests, and the textbook solved this question via the integral test, which leads me to believe that the book has made a mistake as this is not a monotonic decreasing function.

I'd agree with that. You can't really even integrate it in any elementary sense anyway, the integral is Si(x), the sine integral, which doesn't mean much beyond being a formal symbol for the integral. What actually does the textbook say? BTW the Dirichlet test will tell it does converge. Needs a bit of work to show that though.

 

[marcusl]

Thank you for your reply. The text we are using only covers the elementary tests, and the textbook solved this question via the integral test, which leads me to believe that the book has made a mistake as this is not a monotonic decreasing function.

FYI, it's possible to show that the envelope of sinx/x is monotonically decreasing--it decays as 1/x. I will grant, however, that this is not very helpful in solving your problem.

 

[fakecop]

Ok, so I have read about the dirichlet's test. so I can prove that sinx/x converges because 1/x is decreasing and approaching zero, while Ʃ (1 to N) sinx ≤ N for any positive integer N since sinx is bounded above by 1, right?

 

[Dick]

Ok, so I have read about the dirichlet's test. so I can prove that sinx/x converges because 1/x is decreasing and approaching zero, while Ʃ (1 to N) sinx ≤ N for any positive integer N since sinx is bounded above by 1, right?

Not quite. You need to prove that the sine sum has a bound that's independent of N. You can actually derive a formula for the sum and show it's bounded.

 

[fakecop]

I surfed the internet and the only result that does not involve complex exponents is this:

By the sum and difference identities for Cosine, we obtain 2*sin(x)*sin(0.5)=cos(x-0.5)-cos(x+0.5)

When we sum x from 1 to infinity, we get 2*sin(x)*sin(0.5)=cos(0.5)-cos(x+0.5) because there
are mass cancellations on the right side of the equation.

Thus, Ʃ 1 to infinity sin(x) = lim x→infinity (cos(0.5)-cos(x+0.5))/2*sin(0.5), which is bounded above and below because the cosine function is bounded above by 1 and below by -1.

Would this be considered a proof that sinx/x converges?

 

[lurflurf]

first we have
$$\int_0^\infty \! \sin(x)/x \, \mathrm{dx}=\pi/2$$
but the integral test does not require that we know the integral, only that it converges to a finite value.
The monotone issue can be handled by transforming the function

Here is what we said last time someone asked this
I suggested
$$\sum_{n=1}^N \frac{\sin(n \, a)}{n}=-\frac{\cos(N \, a+a/2)}{2(N+1) \sin(a/2)}+\frac{\cot(a/2)}{2}-\sum_{n=1}^N \frac{\cos(n \, a+a/2)}{2n(n+1) \sin(a/2)} $$
The second sum being clearly convergent
or
use
http://en.wikipedia.org/wiki/Abel's_summation_formula

 

[jbunniii]

I'll mention as a side note that if you know a few facts about Fourier transforms, not only can we easily show that $\sum_{n=0}^{\infty} \sin(n)/n$ converges, but we can actually find the value of the sum.

First, recognize that
$$f(x) = \begin{cases}\pi & \text{if }-\frac{1}{2\pi} < x < \frac{1}{2\pi} \\
0 & \text{otherwise}\end{cases}$$
and
$$\hat{f}(y) = \frac{\sin(y )}{y}$$
are a Fourier transform pair, then we can use the Poisson summation formula:
$$\sum_{n = -\infty}^{\infty} f(n) = \sum_{k=-\infty}^{\infty}\hat{f}(k)$$
I'll omit the details because this is probably beyond the scope of the OP's assignment, but once you learn a bit of Fourier analysis, this clearly becomes the "right" way to solve the problem.

 

[mfb]

The integral test needs a monotone decreasing function. f(x)=sin(x)/x does not satisfy that.

 

[STEMucator]

I'll mention as a side note that if you know a few facts about Fourier transforms, not only can we easily show that $\sum_{n=0}^{\infty} \sin(n)/n$ converges, but we can actually find the value of the sum.

First, recognize that
$$f(x) = \begin{cases}\pi & \text{if }-\frac{1}{2\pi} < x < \frac{1}{2\pi} \\
0 & \text{otherwise}\end{cases}$$
and
$$\hat{f}(y) = \frac{\sin(y )}{y}$$
are a Fourier transform pair, then we can use the Poisson summation formula:
$$\sum_{n = -\infty}^{\infty} f(n) = \sum_{k=\infty}^{\infty}\hat{f}(k)$$
I'll omit the details because this is probably beyond the scope of the OP's assignment, but once you learn a bit of Fourier analysis, this clearly becomes the "right" way to solve the problem.

Indeed I believe the second sum should be from $-∞$ to $∞$. I'd prefer this over Dirichlet's test.

 

[jbunniii]

Indeed I believe the second sum should be from $-∞$ to $∞$

Yes, thanks. I'll fix it now.

 

[jackmell]

No way to "force-fit" an "integral-like" test on this series right?

Suppose that was the exercise: force-fit an integral-like test to the series $\sum_{n=1}^{\infty}\frac{\sin(n)}{n}$.

How about I try?

Consider:

$$\frac{u(x)}{x}$$

where $u(x)$ is a uniformly-periodic function with $lim_{x\to 0} \frac{u(x)}{x}=k$ and $u(x)=u(x+a)$ with $u(b)=-u(b+(ka+a/2))$ (like the sine function)

then if:

$$\int_1^{\infty}\frac{u(x)}{x}dx$$

converges then the sum:

$$\sum_1^{\infty} \frac{u(n)}{n}$$

converges as well.

You guys telling me I can't do that? Granted, I need to spruce it up, prove it for one, but it looks reasonable to try.

 

[mfb]

No, you need stronger constraints on u(x). Consider u(x)=cos(2 pi x)...

Personal guess: Assume u is continuous and periodic with an irrational period. I would expect that this is sufficient to let ##\sum \frac{u(n)}{n} converge. If you drop one of those two conditions, I can find a counterexample.

 

[jackmell]

No, you need stronger constraints on u(x). Consider u(x)=cos(2 pi x)...

But $\lim_{x\to 0} \frac{\cos(2\pi x)}{x}=\infty$ so is not sine-like.

Personal guess: Assume u is continuous and periodic with an irrational period. I would expect that this is sufficient to let ##\sum \frac{u(n)}{n} converge. If you drop one of those two conditions, I can find a counterexample.

I have doubts about that: suppose it's periodic but not uniformly so. That is, suppose in one period, the function is primarily positive and only negative for a short interval. I'd rather stipulate the function is uniformly periodic: just as much positive area as negative area in one period.

Can anyone show me a counter example where the sum using such a function would diverge?

Also, I like the irrational period: this introduces essentially a random distribution of positive and negative values in the limit as n goes to infinity (I think).

 

[jbunniii]

Can anyone show me a counter example where the sum using such a function would diverge?

I assume you want a continuous counterexample. Otherwise, I think this works:
$$u(x) = \begin{cases}
\sin(\pi x/2) & \text{ if } 0 \leq x \leq 1 \\
-\sin(\pi x/2) & \text{ if } 1 < x \leq 2 \\
\end{cases}$$
Now extend $u$ periodically outside the interval $[0,2]$.

Then we have $u(n) = 0$ if $n$ is even, and $u(n) = 1$ if $n$ is odd, and $u(x)/x \to 1$ as $x \to 0$. So $\sum_{n=0}^{\infty} u(n)/n = 1 + 1 + \frac{1}{3} + \frac{1}{5} + \cdots = \infty$, but (I think) $\int_{0}^{\infty} u(x)/x$ converges.

[edit] Fixed some typos

 

[jackmell]

I assume you want a continuous counterexample. Otherwise, I think this works:
$$u(x) = \begin{cases}
\sin(\pi x/2) & \text{ if } 0 \leq x \leq 1 \\
-\sin(\pi x/2) & \text{ if } 1 < x \leq 2 \\
\end{cases}$$
Now extend $u$ periodically outside the interval $[0,2]$.

Then we have $u(n) = 0$ if $n$ is even, and $u(n) = 1$ if $n$ is odd, and $u(x)/x \to 1$ as $x \to 0$. So $\sum_{n=0}^{\infty} u(n)/n = 1 + 1 + \frac{1}{3} + \frac{1}{5} + \cdots = \infty$, but (I think) $\int_{0}^{\infty} u(x)/x$ converges.

[edit] Fixed some typos

Hi hbunnii,

But that example has an integer period. I like the irrational period requirement. I think maybe this may turn out to be essential for the proof.

 

[jbunniii]

I think I also have a continuous counterexample, but I'm too lazy to write an explicit piecewise formula.

Suppose we start with the positive part of a single period of sine:
$$s(x) = \begin{cases}
\sin(x) & \text{ if }0 \leq x \leq \pi \\
0 & \text{ otherwise }\\
\end{cases}$$
Now, first scale $s$ so it has width 1 and height 1, and place it so it is centered at $x = 1$. So the support of this part is $[\frac{1}{2}, \frac{3}{2}]$.

Next scale $s$ so it has width $\frac{1}{2}$ and height $-h$, and place one copy of this at $[0, \frac{1}{2}]$ and another at $[\frac{3}{2}, 2]$. Thus we have covered all of $[0,2]$. We choose $h$ so that $\int_{0}^{2} u(x) dx = 0$. (I'll attach a sketch in a minute if I can work out how to do it.)

Then $u$ is continuous, and it has the property that $u(n) = 0$ for even $n$, $u(n) = 1$ for odd $n$, and $u(x)/x \to 1$ as $x \to 0$, just as in the previous example. So again we have $\sum_{n=0}^{\infty}u(n)/n = 1 + 1 + \frac{1}{3} + \frac{1}{5} + \cdots = \infty$, but $\int_{0}^{2}u(x)/x$ converges.

[edit] attached picture, if I did it correctly

 

[jbunniii]

Hi hbunnii,

But that example has an integer period. I like the irrational period requirement. I think maybe this may turn out to be essential for the proof.

Yes, I suspect it is essential. Proving that it is sufficient is another story...

 

[fakecop]

Fourier Analysis is the outside the scope of my understanding. Is there a way to prove that sin(x)/x converges by finding the interval of convergence of its taylor series expansion?

 

[fakecop]

first we have
$$\int_0^\infty \! \sin(x)/x \, \mathrm{dx}=\pi/2$$
but the integral test does not require that we know the integral, only that it converges to a finite value.
The monotone issue can be handled by transforming the function

Here is what we said last time someone asked this
I suggested
$$\sum_{n=1}^N \frac{\sin(n \, a)}{n}=-\frac{\cos(N \, a+a/2)}{2(N+1) \sin(a/2)}+\frac{\cot(a/2)}{2}-\sum_{n=1}^N \frac{\cos(n \, a+a/2)}{2n(n+1) \sin(a/2)} $$
The second sum being clearly convergent
or
use
http://en.wikipedia.org/wiki/Abel's_summation_formula

This may seem very stupid but is $$\int_0^\infty \! \sin(x)/x \, \mathrm{dx}=\pi/2$$ a fact? I can't seem to integrate it with improper integration.
Also, how can you show that $$\sum_{n=1}^N \frac{\sin(n \, a)}{n}=-\frac{\cos(N \, a+a/2)}{2(N+1) \sin(a/2)}+\frac{\cot(a/2)}{2}-\sum_{n=1}^N \frac{\cos(n \, a+a/2)}{2n(n+1) \sin(a/2)} $$? Do you employ sum and difference identities?

 

[jbunniii]

This may seem very stupid but is $$\int_0^\infty \! \sin(x)/x \, \mathrm{dx}=\pi/2$$ a fact? I can't seem to integrate it with improper integration.

It is a fact. I don't think you can show it using elementary calculus. There is no anti-derivative in terms of elementary functions. (Indeed, the function $Si$ was created for this purpose.) You can express $\sin$ as a Taylor series, divide the terms by $x$, and integrate the terms indefinitely to get a series expansion for the anti-derivative. But this won't help you to integrate over $[0,\infty]$ because the Taylor series does not converge uniformly on that interval, and each term integrates to $\infty$.

The only proofs I know of offhand use either Fourier analysis (using the same Fourier transform pair I indicated earlier), or complex analysis.

 

[micromass]

It is a fact. I don't think you can show it using elementary calculus. There is no anti-derivative in terms of elementary functions. (Indeed, the function $Si$ was created for this purpose.) You can express $\sin$ as a Taylor series, divide the terms by $x$, and integrate the terms indefinitely to get a series expansion for the anti-derivative. But this won't help you to integrate over $[0,\infty]$ because the Taylor series does not converge uniformly on that interval, and each term integrates to $\infty$.

The only proofs I know of offhand use either Fourier analysis (using the same Fourier transform pair I indicated earlier), or complex analysis.

There's also a proof that uses differentiation under the integral sign and dominated convergence.

 

[Dick]

The conditional convergence of the integral sin(x)/x has nothing to do with the conditional convergence of the sum of the series sin(n)/n. Everyone knows this, yes?

 

[Dick]

I surfed the internet and the only result that does not involve complex exponents is this:

By the sum and difference identities for Cosine, we obtain 2*sin(x)*sin(0.5)=cos(x-0.5)-cos(x+0.5)

When we sum x from 1 to infinity, we get 2*sin(x)*sin(0.5)=cos(0.5)-cos(x+0.5) because there
are mass cancellations on the right side of the equation.

Thus, Ʃ 1 to infinity sin(x) = lim x→infinity (cos(0.5)-cos(x+0.5))/2*sin(0.5), which is bounded above and below because the cosine function is bounded above by 1 and below by -1.

Would this be considered a proof that sinx/x converges?

That would be a way to prove that the sum sin(n) is bounded. Now if you apply Dirichlet's test, then yes, you can conclude the series sin(n)/n converges.

 

[lurflurf]

This may seem very stupid but is $$\int_0^\infty \! \sin(x)/x \, \mathrm{dx}=\pi/2$$ a fact? I can't seem to integrate it with improper integration.
Also, how can you show that $$\sum_{n=1}^N \frac{\sin(n \, a)}{n}=-\frac{\cos(N \, a+a/2)}{2(N+1) \sin(a/2)}+\frac{\cot(a/2)}{2}-\sum_{n=1}^N \frac{\cos(n \, a+a/2)}{2n(n+1) \sin(a/2)} $$? Do you employ sum and difference identities?

Yes it is a fact here are some amusing ways to show it
http://www.math.harvard.edu/~ctm/home/text/class/harvard/55b/10/html/home/hardy/sinx/sinx.pdf
Yes that identity can be derived either by using trigonometry (difference identities) and algebra, or using summation by parts.
http://en.wikipedia.org/wiki/Summation_by_parts
The identity shows that the given series diverges or converges with another series, which obviously is absolutely convergent.

The conditional convergence of the integral sin(x)/x has nothing to do with the conditional convergence of the sum of the series sin(n)/n. Everyone knows this, yes?

I would not go that far the integral is the limiting case, since for all the cases n=0,1,...
$$ \sum_{k=1}^\infty \sin(k 2^{-n})/k=\pi/2-2^{-(n+1)}$$

 

[mfb]

I have doubts about that: suppose it's periodic but not uniformly so. That is, suppose in one period, the function is primarily positive and only negative for a short interval. I'd rather stipulate the function is uniformly periodic: just as much positive area as negative area in one period.

Oh right, I forgot that requirement.

Also, I like the irrational period: this introduces essentially a random distribution of positive and negative values in the limit as n goes to infinity (I think).

That's the point, and the reason why the series basically "evaluates this integral" (modified with 1/n, which gets arbitrarily close to constant within a period for large n). Without an irrational period, we can pick all values the function will use, find an arbitrary continuous function in between and make the series divergent.