Question

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

Answer 1

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.