Answer

This exercise is the same as the example except that we replaced some of the numbers with letters. The nth partial sum of the series is

S_n = (a₁ + (1 – 1)d) + (a₁ + (2 – 1)d) + ... + (a₁ + (n – 1)d).

This is a sum of n terms. Since both a₁ and d are positive, each term is a positive number greater than or equal to a₁. Thus

S_n ≥ n × a₁.

As n approaches ∞ so does n × a₁, which means

The sequence of partial sums diverges, which means the original arithmetic series diverges.

We made a₁ positive in the exercise to make things easier to think about, but the series would still diverge if a₁ were negative.

If a₁ is negative and d is positive, eventually we'll add enough copies of d that we'll get a positive term

a_n = a₁ + (n – 1)d.

From the exercise we know that, if our initial term and our difference d are both positive, the series diverges. Start the series out at a_n instead of a₁.

The series

has positive initial term a_n and positive difference d, so it diverges.

Since starting at a different term doesn't affect whether the series converges or diverges, the series

has to diverge also.

Our burger analogy makes more sense now. You can dress a burger with special sauce, lettuce, cheese pickles and onions, but it's still a burger. For our arithmetic burger series, it doesn't matter whether a₁ is positive or negative. Either way, the series diverges.

Answer

If a₁ and d are both negative, every term will be negative and at most a₁. In symbols,

a_i ≤ a₁ < 0

for all i. If we sum n such terms, we see that the nth partial sum is

S_n ≤ n(a_i).

As n approaches infinity, the partial sums S_n keep getting more and more negative, so

This means the series diverges.

Just like the previous exercise, it doesn't matter if a₁ is positive or negative. If we keep adding a negative value to a₁, eventually we'll get a negative term a_n. From what we just said, the series

must diverge. Since changing the starting limit of summation doesn't change whether the series diverges,

has to diverge also.

A constant series diverges unless the constant is 0. By comparing arithmetic series to a constant series, we showed that arithmetic series have to diverge unless both a₁ and d are 0.