# Is it possible to evaluate sum_(n=1)^oosqrt(4n^2x^2-1)/(4n^2) in terms of x?

Jun 21, 2018

For $x \ne 0$ let:

${a}_{n} = \frac{\sqrt{4 {n}^{2} {x}^{2} - 1}}{4 {n}^{2}}$

and:

${b}_{n} = \frac{1}{2 n \left\mid x \right\mid}$

Now:

${\lim}_{n \to \infty} {a}_{n} / {b}_{n} = {\lim}_{n \to \infty} 2 n \left\mid x \right\mid \frac{\sqrt{4 {n}^{2} {x}^{2} - 1}}{4 {n}^{2}}$

${\lim}_{n \to \infty} {a}_{n} / {b}_{n} = x {\lim}_{n \to \infty} \frac{\sqrt{4 {n}^{2} {x}^{2} - 1}}{2 n}$

${\lim}_{n \to \infty} {a}_{n} / {b}_{n} = \left\mid x \right\mid {\lim}_{n \to \infty} \sqrt{\frac{4 {n}^{2} {x}^{2} - 1}{4 {n}^{2}}}$

${\lim}_{n \to \infty} {a}_{n} / {b}_{n} = \left\mid x \right\mid {\lim}_{n \to \infty} \sqrt{{x}^{2} - \frac{1}{4 {n}^{2}}}$

${\lim}_{n \to \infty} {a}_{n} / {b}_{n} = {x}^{2}$

Based on the limit comparison test, for any $x \ne 0$ the limit is finite and non null, so that the series must have the same characteristic.

But:

${\sum}_{n = 0}^{\infty} \frac{1}{2 n \left\mid x \right\mid} = \frac{1}{2 \left\mid x \right\mid} {\sum}_{n = 0}^{\infty} \frac{1}{n}$

is divergent, so also:

${\sum}_{n = 0}^{\infty} \frac{\sqrt{4 {n}^{2} {x}^{2} - 1}}{4 {n}^{2}}$

is divergent.