We'll use the integral test to determine the convergence of the given series.

We know that if:

Int f(x)dx, x = 1 to x = infinite,

converges, then the series Sum1/(n^2+17) converges.

We'll determine the integral:

Int [1/(x^2+17)]dx , x = 1 -> x = infinite

If the lim Int [1/(x^2+17)]dx, for x = 1 to x = N, N->infinite, is finite, then the series is convergent.

We'll evaluate first the indefinite integral:

Int [1/(x^2+17)]dx = Int [1/(x^2+(sqrt17)^2)]dx

Int [1/(x^2+(sqrt17)^2)]dx = (1/sqrt17)*arctan(x/sqrt17)

Lim Int [1/(x^2+(sqrt17)^2)]dx = lim (1/sqrt17)*arctan(x/sqrt17)

lim (1/sqrt17)*arctan(x/sqrt17) = lim (1/sqrt17)*arctan(N/sqrt17) - lim (1/sqrt17)*arctan(1/sqrt17), N-> infinite

lim (1/sqrt17)*arctan(x/sqrt17) = pi/2sqrt17 -(1/sqrt17)*arctan(1/sqrt17)

**The given series, where n=1to n = infinite, is convergent.**