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

1 Answer
Jun 21, 2018

For x!=0 let:

a_n = sqrt(4n^2x^2-1)/(4n^2)

and:

b_n = 1/(2nabsx)

Now:

lim_(n->oo) a_n/b_n = lim_(n->oo) 2nabsx sqrt(4n^2x^2-1)/(4n^2)

lim_(n->oo) a_n/b_n = x lim_(n->oo) sqrt(4n^2x^2-1)/(2n)

lim_(n->oo) a_n/b_n = absx lim_(n->oo) sqrt((4n^2x^2-1)/(4n^2))

lim_(n->oo) a_n/b_n = absx lim_(n->oo) sqrt(x^2-1/(4n^2))

lim_(n->oo) a_n/b_n = x^2

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

But:

sum_(n=0)^oo 1/(2nabsx) = 1/(2absx)sum_(n=0)^oo1/n

is divergent, so also:

sum_(n=0)^oo sqrt(4n^2x^2-1)/(4n^2)

is divergent.