As we have seen, the series listed below have the indicated behaviors:

• The harmonic series diverges

• The ratio of its terms approaches 1.

• The p series converges if p > 1 and diverges if p = < 1

• In either case the ratio of its terms approaches 1.

• The geometric series converges if | r | < 1 and diverges if | r | >= 1

• In either case the ratio of its terms is r
• In the convergent case the ratio of its terms is thus < 1 and in the divergent case the ratio of its terms is >= 1.

• The Taylor series about x = a converges within a distance R of a, where R is the reciprocal of the ratio of its coefficients.

• When the ratio of its coefficients has limit 1 / R, then whenever | x - a | < R the ratio of its terms will be be < 1.

Video Clip #01

To determine whether the power series in the figure below converges, we take the limit of the reciprocal ratio of its coefficients:

• Since C_n = 1 / n, the reciprocal ratio C_n / C_n+1 approaches 1.
• The series therefore converges with radius of convergence R = 1.
• Since the series is in powers of (x - 0), it therefore converges for | x - 0 | < 1, or simply | x | < 1.

For the series below, we first attempt to use the ratio test to determine its convergence.

• The ratio test gives us a limiting ratio of 1 for the terms of the series, which is of no help.
• The limit test shows us clearly that the series diverges, since the limit of the terms of the series is not zero.
• The limit test would have been a better choice for the first attempt. Unless the terms of the series are clearly approaching 0, the ratio test will often be useless. In this case the terms of the series were increasing, which clearly indicates divergence; despite this fact the ratio test gave us an inconclusive result.

If we apply the limit test to the series below, we see that the limit is zero and therefore convergence is indeed possible.

• It is important to understand that the limit test cannot assure us that the series converges.

We therefore proceed to use the ratio test.

• The ratio test is again inconclusive, but was easy to use and therefore not such a bad choice.
• We therefore proceed to attempt a comparison test.

• We hypothesize that the series diverges and therefore compare it with a harmonic series.

• If we had rather assumed that the series converges, we would have perhaps used the p series with p = 2 (i.e., 1 / n²).

• We see that the terms of the series are always > those of the divergent harmonic series, and conclude that the series therefore diverges.

Video Clip #02

We use an integral test for the series below, which has nth term n² / (n³ + 10).

• We can arrange rectangles whose areas represent the terms of the series as indicated with the graph of the series.
• Since the integral of x² / (x³+1) diverges, it follows that the series, which 'lies above' the graph of the integral, diverges.

• The actual shape of the graph of the function x² / (x³+1) is shown near the bottom; the shape of the top graph is a generic shape which indicates the behavior of the series beyond the maximum point of the actual curve.

• Note that the finite number of terms that lie to the left of the maximum of the actual curve have no effect on the convergence or divergence of the series, since they correspond to only a finite total.

Video Clip #03

Below we prove that if a₀ + a₁ x + a₁ x² + . . . is convergent with radius of convergence R, then a₀ + 4 a₁ x + 9 a₂ x² + . . . + (n+1)² a_n xⁿ + . . . is also convergent with radius of convergence R.

• The radius of convergence of the second series will be the limit of (n+1)² a_n / [ (n+2)² a_n+1 ].

• This expression can be factored into [ (n+1)² / (n+2)² ] and [ a_n / a_n+1 ].
• The limit of the original expression will therefore be the product of the limits of these factors.

• The first factor has limit 1, while the second has limit R; it is therefore clear that the original expression has limit 1 * R = R.
• The second series therefore has radius of convergence R, which is what we were trying to prove.

In the figure below we prove that if the series with nth term a_n and the series with nth term b_n both converge, then series with nth term (a_n + b_n) must also converge.

• To show the series converges, we will show that the limit of its partial sums is finite.
• Since the a_n series converges, the limit S_n of its partial sums is finite.
• Since the b_n series converges, the limit T_n of its partial sums is finite.
• In the last paragraph we show that the nth partial sum of the (a_n + b_n) series is equal to S_n + T_n.
• Then since S_n and T_n both have finite limits, S_n + T_n has a finite limit.
• Thus the partial sums of the (a_n + b_n) series have a finite limit and the series (a_n + b_n) is therefore convergent.

In the figure below we depict the fact that a series a_n and a series b_n may be as different as we like up to some n = N; if they are identical after that value of n, then they either both convergent or both diverge.

Video Clip #04