Finding a formula for the partial sum *S*_{n} would be a bit annoying, but thankfully we don't need to bother with that. We know that *S*_{n} is the sum of the first *n* terms of the series: *S*_{n} = (4 + (1 – 1)3) + (4 + (2 – 1)3) + ... + (4 + (*n* – 1)3).
Each term is 4 plus something non-negative: *S*_{n} = (4 + (1 – 1)3) + (4 + (2 – 1)3) + ... + (4 + (*n* – 1)3).
Since each of the *n* terms we're adding up is at least 4, the partial sum is at least *n* times 4: *S*_{n} ≥ 4*n*.
From here it should be easy to see that as *n* approaches ∞, so does *S*_{n} (if this isn't easy to see, go review limits). Since is not a finite number, the sequence of partial sums diverges. This means the original arithmetic series diverges. |