1
$\begingroup$

Determine whether the following function is convergent or divergent? If convergent, what does it converge towards.

$$f(x)=-2\sum _{ n=-\infty }^{ \infty }{ \left( \frac { \left( -1 \right) ^{ n } }{ n } \right) } ·sin(nx)$$

Things I have tried

Determining whether it is convergent

So I started by using the Root test to test the convergence and I got 1 for the following expression from the sum,

$$a_n=\left( \frac { \left( -1 \right) ^{ n } }{ n } \right) ·sin(nx)$$

Hence the test is inconclusive. But I have a question on this, because wont I have to time my answer for $a_n$ with -2? Meaning when doing the Test, do I have to write the expression as,

$$-2 \cdot |a_n|^{1/n}$$

or as independent of -2.

$$|a_n|^{1/n}$$.

On the other hand, I also tried using another approach, where I have said that since we know that,

$$\frac{(-1)^n}{n}$$

is conditionally convergent, and therefore our sum should also be conditionally convergent. Is this a correct method? If so, then how can I write that in mathematical term.

The final part of the question, I do not know how can I find the value of the convergence. I mean towards what value does it converge. I mean,$\frac{(-1)^n}{n}$ converges towards $-log(2)$ but how can I find the value of the above function. It is the sin(nx) that has been confusing me. Any help will be great. Thank You.

EDIT The value it converges towards

So I have figured out how to determine whether it is convergent or not. But how can I determine the value it converges towards? Because I get the value of 1, but am sure that is wrong.

$\endgroup$
  • $\begingroup$ +1) The $-2$ has no bearing on whether or not this series is convergent. You are essentially dealing with the alternating harmonic series (which is convergent), whose terms are multiplied by values between $-1$ and $1$. Looks to me that your series should be handled with Comparison Theorem and Sandwich theorem $\endgroup$ – imranfat Nov 12 '16 at 20:49
  • $\begingroup$ So if i use the comparison theorem, then can I use $\frac{(-1)^n}{n}$ or do I have to use some other function? Because $\frac{(-1)^n}{n}$>$\frac{(-1)^n}{n}\cdot sin(nx)$ $\endgroup$ – MathCurious314 Nov 12 '16 at 20:55
  • $\begingroup$ Fact is that the sine term produces values between $-1$ and $1$ and so the "alternating harmonic series" is bounded, i.e. each term of your terms of your series produces values that are LESS than the harmonic series. That is where the sandwich theorem comes in $\endgroup$ – imranfat Nov 12 '16 at 20:57
  • $\begingroup$ If I have not misundestood, so what I have written in my earlier comment is correct. Hence can I simply use that expression to conclude my my series is convergent? $\endgroup$ – MathCurious314 Nov 12 '16 at 20:59
  • $\begingroup$ The factor $\sin (nx)$ can behave rather erratically. One can have $\lvert \sin (kx)\rvert$ small and $\lvert \sin ((k+1)x)\rvert$ close to $1$. That is difficult to control. You probably should use Dirichlet's test. $\endgroup$ – Daniel Fischer Nov 12 '16 at 21:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.