10.6.7 The Ratio Test

Our estimations of `n`th tails of series by comparison to geometric series leads to a consistent procedure that we can apply to any series with infinitely many terms. There are cases in which this procedure will not tell us anything about convergence or divergence, but in many other cases it completely resolves the problem. In particular, this procedure will show us why the interval of convergence of a power series must be symmetric with respect to the reference point, except possibly at the endpoints of the interval.

In all our comparison examples we started by taking absolute values. These comparison techniques work only if all the inequalities go in the same direction. We wouldn't be saying much about the size of an `n`th tail if we said the tail was smaller than something and "something" turned out to be `-10^10`. It's important that all terms of the series being tested are nonnegative.

The effect of always taking absolute values is that we are always dealing with the worst case — no cancellations among the terms being added, so the sum is always getting bigger as we add more terms. If any of the terms of the original series are negative, that can only produce a smaller sum — but not too small, because none of the partial sums can be smaller than the negative of the absolute value series. In symbols,

- ∑ k = 0 ∞ | b k | ≤ ∑ k = 0 ∞ b k ≤ ∑ k = 0 ∞ | b k | .

If the third of these sums is finite, then so are the other two. If the one in the middle happens to have oscillations, they must die out, because convergence of the absolute value series tells us

lim k → ∞ | b k | = 0.

A geometric series has the special property that the ratio of each term to the preceding term is constant. That is, the `k`th term in `1+r+r^2+r^3+cdots` is `r^k`, and `b_(k+1)  text[/] b_k=r^(k+1)  text[/] r^k=r` for every `k`. The series converges when `text[|] r text[|] <1` and diverges otherwise. Thus we can expect a successful comparison to a convergent geometric series if the terms in the series under consideration satisfy

| b k + 1 b k | ≤ r

for some constant `r<1` and for all sufficiently large values of `k`. In particular, that will be the case if

lim k → ∞     | b k + 1 b k |

is a number less than `1`. We can then take `r` to be a slightly larger number — still less than `1` — and our ratios will eventually be that small. We illustrate this idea with the series we have already compared to geometric series.

Activity 4

1. For the exponential series,

   1 + x + 1 2 ⁢ x 2 + 1 6 ⁢ x 3 + ⋅⋅⋅ + 1 k ! ⁢ x k + 1 ( k + 1 ) ! ⁢ x k + 1 + ⋅⋅⋅ ,

   calculate `|b_(k+1)//b_k|` and simplify. How is this ratio related to the `r` we found in estimating the size of the `n`th tail in Example 2?
2. Find

   lim k → ∞     | b k + 1 b k |

   for this series. How do you know that `|b_(k+1)//b_k|` must eventually be smaller than some constant less than `1`?
3. For the logarithmic series,

   x - 1 2 ⁢ x 2 + 1 3 ⁢ x 3 - 1 4 ⁢ x 4 + ⋅⋅⋅ ,

   calculate `|b_(k+1)//b_k|` and simplify.
4. Find

   lim k → ∞     | b k + 1 b k |

   for this series. For what values of `x` will `|b_(k+1)//b_k|` eventually be smaller than some constant less than `1`? How do you know?

Comment on Activity 4

The calculations for the exponential and logarithmic series are superficially similar, but the results are quite different. In the first case, we get a limit of ratios that is `0` no matter what `text[|] x text[|]` is — so we get a convergent series for every value of `x`. In the second case, the limiting value depends on `x` — so we get convergence only if `x` is small enough. In both cases, however, we can say that the interval in which the absolute value series converges has the form `-a<x<a`, provided we allow the possibility that `a=oo`.

Activity 5

1. What can you say about convergence or divergence of a series `sum b_k` if

   lim k → ∞     | b k + 1 b k | > 1 ⁢ ?

2. What is

   lim k → ∞     | b k + 1 b k |

   for the harmonic series

   1 + 1 2 ⁢ + 1 3 ⁢ + 1 4 ⁢ + ⋅⋅⋅ + 1 k + ⋅⋅⋅ ?

3. What is

   lim k → ∞     | b k + 1 b k |

   for the alternating harmonic series

   1 - 1 2 ⁢ + 1 3 ⁢ - 1 4 ⁢ + ⋅⋅⋅ + ( - 1 ) k + 1 k + ⋅⋅⋅ ?

4. What can you say about convergence or divergence of a series `sum b_k` if

   lim k → ∞     | b k + 1 b k | = 1 ⁢ ?

Comment on Activity 5

We summarize our observations in the last two Activities in the Ratio Test. We assume for the given series `sum b_k` that `lim_(k rarr oo) |b_(k+1)//b_k|` exists.

Ratio Test   Let L = lim k → ∞     | b k + 1 b k | . If `L<1`, then the series converges. If `L>1`, then the series diverges. If `L=1`, then this test gives no information about convergence or divergence.

Checkpoint 6

Next we explore what the Ratio Test tells us about convergence of power series in terms of the coefficients of those series. Suppose we have a power series `sum_(k=0)^oo c_k x^k` for which all the coefficients are different from `0` and `lim_(k rarr oo) |c_(k+1)//c_k|` exists. Then

| b k + 1 b k | = | c k + 1   x k + 1 c k   x k | =   | c k + 1 c k | ⁢   | x |   ,

and

L = lim k → ∞     | b k + 1 b k | = | x | lim k → ∞     | c k + 1 c k |   .

Let's write `R` for the reciprocal of the limit on the right:

R =   1 lim k → ∞ | c k + 1 c k | ⁢   .

If the limit of ratios of consecutive coefficients happens to be `0`, we let `R=oo.` Then `L= text[|] x text[|] text[/] R`, and `L<1` exactly when `text[|] x text[|] <R`. If `R` happens to be `oo` this means `L<1` for every `x`. In fact, `L=0` for every `x` and `0` is certainly less than `1`.

Thus the Ratio Test tells us that the series converges for every `x` in the symmetrical interval `-R<x<R`. For `x=+-R`, we get no information. Those are the values of `x` for which we have to resort to other tests of convergence. For example, if the power series happens to have alternating signs at an endpoint, we can try the Alternating Series Test.

Ratio Test for Power Series   Suppose that `sum_(k=0)^oo c_k x^k` is a power series with all coefficients `c_k` different from `0` and that `lim_(k rarr oo) text[|] c_(k+1) text[/] c_k text[|]` exists. Let R =   1 lim k → ∞ ⁢ | c k + 1 c k | ⁢   . Then the series converges for `-R<x<R` and diverges for `text[|] x text[|] >R.` The test gives no information about convergence or divergence for `x=+-R`.

Suppose that `R=1` for a particular power series. Then we know that the series converges for each in the open interval `text[(]-1,1text[)]`. Without more information, we cannot say whether the series converges at either or both of the endpoints, `-1` and `1`. The collection of `x` for which the series converges will be one of the following four intervals: `text[(]-1,1text[)]`, `text([)-1,1text[)]`, `text[(]-1,1text(])`, or `text([)-1,1text(])`.

Definitions   If `sum_(k=0)^oo c_k x^k`, is a power series with all coefficients `c_k` different from `0`, if `lim_(k rarr oo) text[|] c_(k+1) text[/] c_k text[|]` exists, and if R =   1 lim k → ∞ ⁢ | c k + 1 c k | ⁢   , then `R` is called the radius of convergence of the series, and the interval `text[(]-R,Rtext[)]` is called the open interval of convergence of the series.

Example 5   Find the open interval of convergence of the series

∑ k = 0 ∞ ( - 1 ) k x k ( k + 1 )   2 k ⁢   .

What can be said about convergence at the endpoints of the interval?

Solution   We first calculate the limit of absolute values of ratios of consecutive coefficients:

lim k → ∞ ⁢ | c k + 1 c k | = lim k → ∞ 1 / | ( k + 2 )   2 k + 1 | 1 / | ( k + 1 )   2 k | ⁢ = lim k → ∞   k + 1 ( k + 2 )   2  = 1 2 .

The radius of convergence, therefore, is `2`, and the series converges for `-2<x<2`. At `x=2`, the series is

    ∑ k = 0 ∞ ( - 1 ) k k + 1 ⁢   ,

which is the alternating harmonic series. We know this series converges — in fact, we know its sum is `ln 2`. At `x=-2` the series is

    ∑ k = 0 ∞ 1 k + 1 ⁢   ,

the harmonic series, which we know diverges.

Conclusion: The series converges for `-2<x<=2` and diverges otherwise.

Checkpoint 7

So far we have considered the Ratio Test only for power series in which every power of `x` has a nonzero coefficient. The general Ratio Test works for other power series as well, as we illustrate in the next example. We will pursue this further in the exercises.

Example 6   Find the open interval of convergence of the series

∑ k = 0 ∞ ( - 1 ) k   x 2 k + 1 ( k + 1 )   2 k ⁢ .

What can be said about convergence at the endpoints of the interval?

Solution   This series has only odd-power terms. It starts out

x - x 3 2 ⋅ 2 + x 5 3 ⋅ 2 2 - x 7 4 ⋅ 2 3 + ⋅⋅⋅ ⁢ .

Using the Ratio Test for series in general, we find

lim k → ∞     | b k + 1 b k | = lim k → ∞     | x | 2 k + 3 / | ( k + 2 )   2 k + 1 | | x | 2 k + 1 / | ( k + 1 )   2 k | = lim k → ∞ | x | 2 ( k + 2 )   2 = | x | 2 2 .

This limit will be less than `1` when `|x|^2<2`, that is, for `-sqrt(2)<x<sqrt(2)`. At `x=sqrt(2)` we have `x^(2k+1)=sqrt(2)^(2k+1)=2^ksqrt(2)`. Thus the series becomes

∑ k = 0 ∞ ( - 1 ) k 2 k + 1 ,

or `sqrt(2)` times the alternating harmonic series. We know this converges, and in fact the sum is `sqrt(2) ln 2~~0.98`. At `x=-sqrt(2)` the only difference is a sign change — this series has odd symmetry — so we have convergence to `-sqrt(2) ln 2`.

Conclusion: The series converges for `-sqrt(2)<=x<=sqrt(2)`.

We see in Example 6 that the effect of having only odd-power terms in a power series is to introduce a factor of `|x|^2` in the limit of ratios, which leads to a square root in the calculation of endpoints of the open interval of convergence. The same thing happens if there are only even-power terms. We could work out special ratio tests for these cases, but it is simpler to refer back to the general Ratio Test rather than have a plethora of special tests.

Finally, we note that a radius of convergence exists for all power series `sum c_kx^k`, whether or not `lim_(k rarr oo) text[|] c_(k+1) text[/] c_k text[|]` exists. However, for all the power series you will encounter in this course, the limit of the absolute values of ratios of successive coefficients does exist — possibly allowing for regularly occurring zero coefficients, as we illustrated in Example 6.