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.