Answer

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

*S*_{n} = (*a*_{1} + (1 – 1)*d*) + (*a*_{1} + (2 – 1)*d*) + ... + (*a*_{1} + (*n* – 1)*d*).

This is a sum of *n* terms. Since both *a*_{1} and *d* are positive, each term is a positive number greater than or equal to *a*_{1}. Thus

*S*_{n} ≥ *n* × *a*_{1}.

As *n* approaches ∞ so does *n* × *a*_{1}, which means

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

We made *a*_{1} positive in the exercise to make things easier to think about, but the series would still diverge if *a*_{1} were negative.

If *a*_{1} is negative and *d* is positive, eventually we'll add enough copies of *d* that we'll get a positive term

*a*_{n} = *a*_{1} + (*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 *a*_{1}.

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 *a*_{1} is positive or negative. Either way, the series diverges.