I've been given a question:

Let pn denote the nth prime number. Then p1 = 2, p2 = 3, p3 = 5, p4 = 7, p5 = 11, . . . .

It is known that the infinite sum 1/p1 + 1/p2 + 1/p3 + · · · + 1/pn + · · · = infinity.

Find the smallest positive integer N so that 1/p1 + 1/p2 +1/p3 + · · · + 1/pN−1 + 1/pN > e. [Hint : ithprime(n) generates the nth prime number.]

How do I start off?

Many thanks!

The question is to solve

x^2*(diff(y(x), x, x))+x^3*(diff(y(x), x))+(x^2-2)*y(x) = 0

and evaluate it where y(1) = 1 and y(2) = 2 to find y(3).

When I do the equation,

ODE := x^2*(diff(y(x), x, x))+x^3*(diff(y(x), x))+(x^2-2)*y(x) = 0

dsolve(ODE) y(x) = _C1/x+_C2*(-sqrt(Pi)*sqrt(2)*erf((1/2)*sqrt(2)*x)+2*x*exp(-(1/2)*x^2))/x

I have gotten an erf. Is this correct?