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
Sn ≥ 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
an = 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 an instead of a1.
has positive initial term an 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.
Look at the arithmetic series
ai = a1 + (i – 1)d
and d is negative. Does this series converge or diverge?
If a1 and d are both negative, every term will be negative and at most a1. In symbols,
ai ≤ a1 < 0
for all i. If we sum n such terms, we see that the nth partial sum is
Sn ≤ n(ai).
As n approaches infinity, the partial sums Sn 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 an. 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.