How can I deal with this series?

[Hernaner28]

Homework Statement

\displaystyle \sum\limits_{n=1}^{\infty }{\frac{\sin \left( \frac{1}{n} \right)-\frac{1}{n}}{{{n}^{\alpha }}}}

Homework Equations

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!

 

[Mute]

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 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.

 

[Hernaner28]

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

 

[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?

 

[Hernaner28]

Is it?

 

[Aero51]

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

 

[Hernaner28]

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

 

[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.

 

[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!

 

[Mute]

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?

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.

 

[Hernaner28]

Thank you!