Here we will prove that if d > 0, then the arithmetic sequence a_n = a + nd is not bounded from above.

Consider an arbitrary real number K. We will show that some elements of the sequence are larger than K, so it cannot be an upper bound. In other words, we need to find some n satisfying a_n > K. This reads a + nd > K, that is, n > (K − a)/d. Since on the right we have a concrete real number, for sure there is some natural number n satisfying this inequality, but then it also satisfies a_n = a + nd > K, exactly as claimed.

In fact, note that given a real number K, we can choose one particular natural number N satisfying N > (K − a)/d and then for all n = N, N + 1, N + 2,... we have a_n > K. This proves by definition that if a > 0, then the arithmetic sequence a_n = a + nd goes to infinity.