Evaluate the sum of series of arctan and find limit

limitssummationtrigonometric series

I try to find the value of following limit:

$$
\lim _{n \rightarrow \infty} 2^{-n/2} \sum_{j=2^{n}+1}^{2^{n+1}} \tan ^{-1}\left(\frac{1}{\sqrt{j-1}}\right)
$$

But I cannot evaluate the summation. Can anybody help?

Best Answer

METHODOLOGY $1$: Simple Bounds and Using the Squeeze Theorem

Since the arctangent is monotonically increasing, $\arctan(1/\sqrt{x-1})$ is monotonically decreasing. Hence we can assert that

$$\int_{N}^{M+1}\arctan\left(\frac1{\sqrt{x-1}}\right)\,dx\le \sum_{j=N}^M\arctan\left(\frac1{\sqrt{j-1}}\right)\le \int_{N-1}^{M}\arctan\left(\frac1{\sqrt{x-1}}\right)\,dx\tag1$$

The antiderivative of $\arctan\left(\frac1{\sqrt{x-1}}\right)$ can be written as

$$\int \arctan\left(\frac1{\sqrt{x-1}}\right)\,dx=\sqrt{x-1}+x\arctan\left(\frac1{\sqrt{x-1}}\right)+C\tag2$$

For $N=2^n+1$ and $M=2^{n+1}$ in $(2)$, we find that

$$\begin{align} \int_{2^{n}+1}^{2^{n+1}+1} \arctan\left(\frac1{\sqrt{x-1}}\right)\,dx&=2^{n/2}(\sqrt 2-1)\\\\ &+(2^{n+1}+1)\arctan\left(\frac1{\sqrt{2}2^{n/2}}\right)\\\\ &-(2^n+1)\arctan\left(\frac1{2^{n/2}}\right)\tag3 \end{align}$$

Dividing $(3)$ by $2^{n/2}$ and letting $n\to\infty$ reveals that

$$\liminf_{n\to\infty}2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)\ge 2(\sqrt 2-1)\tag4$$

Similarly, evaluation of the right-hand side of $(1)$ reveals

$$\limsup_{n\to\infty}2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)\le 2(\sqrt 2-1)\tag5$$

Finally, putting $(4)$ and $(5)$ together yields the coveted limit

$$\lim_{n\to\infty} 2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)=2(\sqrt 2-1)$$

METHODOLOGY $2$: Use on the Euler-Maclaurin Summation Formula

From the Euler-Maclaurin Summation formula we have

$$\begin{align} 2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)&=2^{-n/2}\int_{2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{x-1}}\right)\,dx+o(1)\\\\ &=2^{-n/2}\left(\sqrt{2^{n+1}-1}-\sqrt{2^n} \right)\\\\ &+2^{n/2+1}\arctan\left(\frac1{\sqrt{2^{n+1}-1}}\right)\\\\ &-(2^{n/2}+1)\arctan\left(\frac1{\sqrt{2^{n}}}\right)+o(1)\tag6 \end{align}$$

Letting $n\to \infty$ in $(6)$, we find that

$$\lim_{n\to\infty} 2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)=2(\sqrt 2-1)$$

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