Solving Sequence: Prove Decreasing with n

[dextercioby]

So I was trying to show that

$$a_n = n \tan\frac{\pi}{n}$$

is decreasing as n increases (n>=3) . I can't see it. Anyone may help ? :)

 

[daveb]

tan x = x + 1/3x³ + 2/15x⁵ + 17/315x⁷ + ...for |x| < π/2, so expand a_n and subtract from it a_n+1 and you should see this

 

[dextercioby]

I was aiming at a method from first principles (tangent is the ratio of sine and cosine), without derivatives, because I know that I can take n from N (>=3) to the real subset [3,infty) and analyze the sign of the first derivative of the function. It's negative for x>0, hence the function keeps decreasing as x increases. Then the monotonicity of the sequence follows.

Your definition of tangent needs to be proven equivalent to the triangle one.

 

[Millennial]

I tried the tangent difference identities and such but it does not help. I think it is best to use the sine-tangent inequality to evaluate the derivative of sine and cosine, and then using them to evaluate the derivative of tangent. I doubt there is any better way using elementary means.

 

[dextercioby]

Going further, the proof I'm searching reduces to show that once $\displaystyle{n\in \mathbb{N}}$, the sequence

$$a_n = n \sin\frac{1}{n}$$

increases with n.

How do I prove it ?

 

[Infinitum]

Going further, the proof I'm searching reduces to show that once $\displaystyle{n\in \mathbb{N}}$, the sequence

$$a_n = n \sin\frac{1}{n}$$

increases with n.

How do I prove it ?

I'm not sure about this, but here's what I thought.

You can try seeing whether n increases faster than sin(1/n) decreases, by taking their derivatives(or some other method). You'll reach to the conclusion that n does increase at a faster rate than sin(1/n) decreases, hence, your original function increases.

 

[Millennial]

I think you did not understand his point. We can easily differentiate the function and conclude that it increases everywhere because it has a positive derivative. He wants a proof that includes only trigonometric identities, and I am sceptic about whether it is possible to present such a proof.

I think that the first thing to do would be to express one of the sines in terms of the other one. To do this, I used this course of action:
$$\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}$$
Then, I conclude that $\frac{1}{n+1}+\frac{1}{n(n+1)}=\frac{1}{n}$. I now tried to apply this to the sine sum formula as follows:
$$\sin\left(\frac{1}{n}\right)=\sin\left(\frac{1}{n+1}+\frac{1}{n(n+1)}\right)$$
$$=\sin\left(\frac{1}{n+1}\right)\cos\left(\frac{1}{n(n+1)}\right)+$$
$$\cos\left(\frac{1}{n+1}\right)\sin\left(\frac{1}{n(n+1)}\right)$$
I then rearranged this into the inequality
$$n\sin\left(\frac{1}{n+1}\right)\cos\left(\frac{1}{n(n+1)}\right)+n\cos\left(\frac{1}{n+1}\right)\sin\left(\frac{1}{n(n+1)}\right) \leq n\sin\left(\frac{1}{n+1}\right)+\sin\left(\frac{1}{n+1}\right)$$
which can be rewritten as
$$n\cos\left(\frac{1}{n(n+1)}\right)+n\cot\left(1/(n+1)\right)\sin\left(\frac{1}{n(n+1)}\right)\leq n+1$$
It is straightforward to show that the first term in the LHS is less than or equal to n, so it appears we only need to prove
$$\cot\left(\frac{1}{n+1}\right)\sin\left(\frac{1}{n(n+1)}\right)\leq 1/n$$
I am stuck here. This inequality is equivalent to the above function increasing monotonically.

Can anyone carry this on?

Note: After graphing the function using a program, I am convinced that the inequality holds when n is greater than one, and even more: 1/n and this function is asymptotic, their quotient has a limit of one.

 

[Infinitum]

$$\cot\left(\frac{1}{n+1}\right)\sin\left(\frac{1}{n(n+1)}\right)\leq 1/n$$
I am stuck here. This inequality is equivalent to the above function increasing monotonically.

Can anyone carry this on?

Here's what I did,

We know that sin(x) < x, for all x.

$$sin\left ( \frac{1}{n(n+1)} \right ) < \frac{1}{n(n+1)}$$

Substituting in the inequality Millennial had,

$$cot\left ( \frac{1}{n+1} \right ) \cdot \frac{1}{n(n+1)} \leq \frac{1}{n} \\<br /> <br /> cot\left ( \frac{1}{n+1} \right ) \cdot \frac{1}{n+1} \leq 1$$Having n>2 it follows that,

$$tan\left ( \frac{1}{n+1} \right ) \cdot (n+1) \geq 1$$

We know that tan(x) > x for any x. Hence at the least possible condition, let us assume tan(x) = x. So we get,

$$\left ( \frac{1}{n+1} \right )\cdot (n+1) \geq 1$$

Hence the proof.

Please tell me if there is a gap in my logic, as this I've not much practice with such questions

 

[Millennial]

Here's what I did,

We know that sin(x) < x, for all x.

$$sin\left ( \frac{1}{n(n+1)} \right ) < \frac{1}{n(n+1)}$$

Substituting in the inequality Millennial had,

$$cot\left ( \frac{1}{n+1} \right ) \cdot \frac{1}{n(n+1)} \leq \frac{1}{n} \\<br /> <br /> cot\left ( \frac{1}{n+1} \right ) \cdot \frac{1}{n+1} \leq 1$$


Having n>2 it follows that,

$$tan\left ( \frac{1}{n+1} \right ) \cdot (n+1) \geq 1$$

We know that tan(x) > x for any x. Hence at the least possible condition, let us assume tan(x) = x. So we get,

$$\left ( \frac{1}{n+1} \right )\cdot (n+1) \geq 1$$

Hence the proof.

Please tell me if there is a gap in my logic, as this I've not much practice with such questions

Yes, that is quite elegant. I did not see that, probably due to me being busy with other questions. Thanks for your help with my proof :)

 

[Infinitum]

Yes, that is quite elegant. I did not see that, probably due to me being busy with other questions. Thanks for your help with my proof :)

Yay! I was solving some integrals too, this was a fun break

Your method to get that inequality was really insightful. Probably wouldn't ever have thought of manipulating it that way. Thanks for the idea!