Can the Taylor expansion be used to find the limit of the sequence?

Question

Can we use the Taylor expansion method to find the limit of this sequence, like the following solution?

$ \lim _{n\to \infty }\left(n^2\left(arctan\left(\frac{a}{n}\right)-arctan\left(\frac{a}{n+1}\right)\right)\right)$

a is a constant number.

my solution

Answer 1

The solution in your linked picture is correct (hoping that you truly understand what you are doing with the $o$-expressions). But there is no need to turn the multiplication by $n^2$ into a division by $\frac 1{n^2}$ which only complicates the expression unnecessarily.

You can also calculate the limit using the mean value theorem:

• $\arctan x - \arctan y =\frac 1{1+\xi^2}(x-y)$ with $\xi$ between $x$ and $y$

Using this, you can squeeze the expression.

For $a>0$ you have

$$\frac{1}{1+\left(\frac an\right)^2}\left(\frac an - \frac a{n+1}\right)\leq \arctan \frac an - \arctan \frac a{n+1} \leq \frac{1}{1+\left(\frac a{n+1}\right)^2}\left(\frac an - \frac a{n+1}\right)$$

Hence, multiplying by $n^2$ you get

$$\underbrace{\frac{n^2}{n^2+n}\frac{a}{1+\left(\frac an\right)^2}}_{\stackrel{n\to\infty}{\longrightarrow}\boxed{a}} \leq n^2(\arctan \frac an - \arctan \frac a{n+1} ) \leq \underbrace{\frac{n^2}{n^2+n}\frac{a}{1+\left(\frac a{n+1}\right)^2}}_{\stackrel{n\to\infty}{\longrightarrow}\boxed{a}}$$

For $a<0$ you only have to reverse the relation signs.

Answer 2

Start writing $$\arctan\left(\frac{a}{n}\right)-\arctan\left(\frac{a}{n+1}\right)=\arctan\left(\frac{a}{n^2+n+a^2}\right)$$When $n$ is large, we have $$\arctan\left(\frac{a}{n^2+n+a^2}\right)\sim \arctan\left(\frac{a}{n^2}\right)\sim \frac{a}{n^2}$$ No much need of Taylor series except if you want to go beyond the limit.

In such a case, composition of Taylor series would give $$\arctan\left(\frac{a}{n}\right)-\arctan\left(\frac{a}{n+1}\right)=\frac{a}{n^2}-\frac{a}{n^3}+O\left(\frac{1}{n^4}\right)$$ which will show the limit and how it is approached.