For what value of i does while (i == i + 1) {} loop forever?

double i = Double.POSITIVE_INFINITY;

These puzzles are described in detail in the "Java Puzzlers: Traps, Pitfalls, and Corner Cases" book by Joshua Bloch and Neal Gafter.

double i = Double.POSITIVE_INFINITY;
while (i == i + 1) {}

or:

double i = 1.0e40;
while (i == i + 1) {}

both will result in an infinite loop, because adding 1 to a floating-point value that is sufficiently large will not change the value, because it doesn't "bridge the gap" to its successor¹.

A note about the second puzzle (for future readers):

double i = Double.NaN;
while (i != i) {}

also results in an infinite loop, because NaN is not equal to any floating-point value, including itself ².

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^{1 - Java Puzzlers: Traps, Pitfalls, and Corner Cases (chapter 4 - Loopy Puzzlers).}

^{2 - JLS §15.21.1}

First of all, since the while (i == i + 1) {} loop doesn't change the value of i, making this loop infinite is equivalent to choosing a value of i that satisfies i == i + 1.

There are many such values:

Let's start with the "exotic" ones:

double i = Double.POSITIVE_INFINITY;

or

double i =  Double.NEGATIVE_INFINITY;

The reason for these values satisfying i == i + 1 is stated in
JLS 15.18.2. Additive Operators (+ and -) for Numeric Types:

The sum of an infinity and a finite value is equal to the infinite operand.

This is not surprising, since adding a finite value to an infinite value should result in an infinite value.

That said, most of the values of i that satisfy i == i + 1 are simply large double (or float) values:

For example:

double i = Double.MAX_VALUE;

or

double i = 1000000000000000000.0;

or

float i = 1000000000000000000.0f;

The double and float types have limited precision, so if you take a large enough double or float value, adding 1 to it will result in the same value.