# How can I deal with this series?

1. Jul 2, 2012

### Hernaner28

1. The problem statement, all variables and given/known data
$$\displaystyle \sum\limits_{n=1}^{\infty }{\frac{\sin \left( \frac{1}{n} \right)-\frac{1}{n}}{{{n}^{\alpha }}}}$$

2. Relevant equations

3. The attempt at a solution

I've applied first the nth-test and I verified that the sequence tends to zero.

Then I realized that sin(1/n) is always positive, since the first term is 1, and sine is positive between 0 and 1. So no need to use alternate and absolute convergence tests.

Now I tried to apply the limit comparison test but I cannot find an equivalent function as n tends to infinity. This is not working don't know why:
We have that
$$\displaystyle \sin \left( \frac{1}{n} \right)\approx \frac{1}{n}$$
so
$$\displaystyle \frac{\sin \left( \frac{1}{n} \right)-\frac{1}{n}}{{{n}^{\alpha }}}\approx \frac{1}{{{n}^{\alpha +1}}}$$

What should I do? Ratio test and root test are useless...

Thanks!

2. Jul 2, 2012

### Mute

What you've written here isn't correct. The -1/n cancels the +1/n from the expansion of the sine. The leading order term in the numerator is -(1/3!)(1/n)^3.

If you want bounding sums, consider what the possible maximum and minimum values of the numerator are.

3. Jul 2, 2012

### Hernaner28

Thank you! I will try it and post if I get something

4. Jul 2, 2012

### Hernaner28

ALright so:

$$\displaystyle 0\ge \frac{\sin \left( \frac{1}{n} \right)-\frac{1}{n}}{{{n}^{\alpha }}}\ge \frac{-1}{{{n}^{\alpha }}}$$

So the series should converge for all aplha greater or equal to 2. Is it right now?

5. Jul 2, 2012

Is it?

6. Jul 2, 2012

### Aero51

Why not try a Taylor series expansion of the sine function then calculate the convergence?

7. Jul 2, 2012

### Hernaner28

I only know how to do a Taylor expansion at a concrete point, not to infinity... how is that defined?

8. Jul 2, 2012

### Aero51

The TSE of sine is
$sin(x) ≈ x - x^{3}/3! +x^{5}/n! - x^{7}/7! +...$
substituting in 1/n for x, it is clear that the higher order terms tend to 0 much faster than the first term as n approaches infinity. This justifies the original post that sin(1/n)≈ 1/n as n approaches some large value. And I believe your function will become 0 if this is case. As far as I can remember this approach should be valid but I have not taken Calc III in a few years.

9. Jul 2, 2012

### Hernaner28

This is Calculus I. What you've wrote is the TSE of sine at zero. That doesn't add anything, I've already used that sin(1/n) is equivalent to 1/n when n tends to infity

Thanks!

Last edited: Jul 2, 2012
10. Jul 3, 2012

### Mute

I haven't double checked those bounds, but assuming they're correct, that should work. What I had in mind was

$$-1 \leq \sin\left(\frac{1}{n}\right) \leq 1,$$

and so

$$-1 - \frac{1}{n} \leq \sin\left(\frac{1}{n}\right) \leq 1 - \frac{1}{n}.$$

Since 1 - 1/n is also less than one, it shows that as long as alpha is greater than 1, the series should converge since the sum over 1/n^alpha will converge for alpha > 1 (or, more precisely, any complex number alpha with real part greater than 1).

So, you can be pretty sure that the sum converges, at least for alpha > 1.

I suppose this doesn't strictly guarantee that the sum won't converge for some lower value of alpha - we might be able to find a smaller upper bound, after all. Using the Taylor series expansion as you've already tried, you may be able to come up a revised bound for alpha. For instance, since for large n, sin(1/n) - 1/n ~ -1/n^3 to leading order, you might be able to show that since $( \sin(1/n) - 1/n)/n^\alpha \sim -(1/3!)(1/n^{3+\alpha})$ that the sum should converge for alpha > -2. You just need to try and make that argument rigorous.

11. Jul 3, 2012

Thank you!