I have the following sequence and I want to prove the monotonicity of this sequence

Question

The following sequence is provided and I am required to prove that the sequence is monotone and convergent.

$\left \{ \cos(\frac {1}{2}\arctan (\frac{-n}{2})^n)\right \}$

As far as convergence of the sequence is concerned, it will surely converge as for large $n$, the $\arctan (\frac{-n}{2} )^n$ part will tend to $π/2$. Consequently, the whole sequence will converge to $\frac{1}{√2}$. But, I am not able to prove the monotonicity of the given sequence. Any help would be appreciated. Thanks in advance.

Answer 1

First, observe that the values of the argument of the arctangent, namely $\;\left(-\frac n2\right)^n\;$ , are going "to jump" from positive to negative values (according to $\;n\;$ even or odd, of course), but we don't care as $\;\cos x\;$ is an even function. Second, we have that

$$-\frac\pi2<\arctan x<\frac\pi2\implies-\frac\pi4<\frac12\arctan x<\frac\pi4\;$$

so all the values of $\;\frac12\arctan\left(-\frac n2\right)^n\;$ are within the interval $\;\left[-\frac\pi4,\,\frac\pi4\right]\;$ . Thus, we can write

$$\cos\left[\frac12\arctan\left(\frac{-n}2\right)^n\right]=\cos\left[\frac12\arctan\left(\frac n2\right)^n\right]$$

and also: as $\;n\to\infty\;$ , we have that

$$\;\left(\frac n2\right)^n\xrightarrow[n\to\infty]{}\infty\implies\frac12\arctan\left(\frac n2\right)^2\nearrow{}\frac\pi 4\implies\cos\left[\frac12\arctan\left(\frac n2\right)^n\right]\searrow\;\ldots$$

and you then get your sequence is monotone descending and obviously (?) bounded below, so it is convergent (in fact, the above already gives you the value of the limit, if you payed attention...)

Answer 2

Thanks to the cosine, you can ignore the minus sign. Then $(\frac n2)^n$ is obviously a growing function for $n\ge1$. The arctangent is strictly growing on $\mathbb R$ and outputs a value in $[0,\frac\pi2)$. Finally, the cosine is decreasing in $[0,\frac\pi4)$, and bounded below.

The sequence very quickly tends to $\sqrt{\frac12}$.