Calculus Of One Real Variable – By Pheng Kim Ving
1. The Root Test (RooT)
This section discusses two more convergence tests: the Root
Test and the Ratio Test. First we prove a limit that'll be used later
on in this section.
The Limit Of x1/x As x Approaches Infinity
For real variable x:
The Root Test
If the mth root of x, where m is large, is k, where 0 < k < 1, then x, which is the mth power of k, must be very small.
Theorem 1.1 – The Root Test (RooT)
Note that the proof of the RooT makes use of the SCT.
Test the convergence of the series:
The RooT, which tests the nth root of the general term, is especially useful when the general term is an nth power.
Determine the convergence/divergence of these series using the RooT:
In part a, the fraction in brackets will approach 2/3 and 0
< 2/3 < 1, so for large n it behaves
like the convergent geometric
series of (2/3)n, thus we expect to determine that the series converges. Similarly for part b, where the series diverges to
infinity because 3/2 > 1. We can utilize this limit 3/2 > 1 to declare that the series diverges to infinity, but the problem
statement asks us to use the RooT, hence we must use it.
2. The Ratio Test (RaT)
Consider the geometric series:
The examination of the limit of the ratio an+1/an of the terms of a
series in an attempt to determine the convergence or
divergence of the series is known as the ratio test, abbreviated as “RaT”.
Theorem 2.1 – The Ratio Test (RaT)
The two series
have the same ratio limit of 1, but one converges while the other diverges.
This completes the proof.
The conclusion of the RaT is the same as that of the RooT. Remark that the proof of the RaT makes use of the SCT.
Test each of the following series for convergence:
a. Let an = n1,000/3n. Then:
Note that we use the relation (n
+ 1)! = n!(n + 1) (since (n + 1)! = 1 x 2 x 3 x ... x n x (n + 1) = (1 x 2 x 3 x ... x n)
x (n + 1)).
The RaT is particularly useful when an exponential involving n and/or a factorial involving n are present in the general term.
Remark that n1,000
contributes a factor equivalent to 1 (= 11,000) in the expression of the last
limit, 3n contributes a factor 3, and
n! contributes a factor n + 1. We can use this aspect of contribution to skip the intermediate steps, as follows:
Relative Speeds Of Increase
Part a involves a power of n
(n1,000) and an exponential in n (3n). Part b involves an
exponential and a factorial. Since both
series converge, the limits of n1,000/3n and 3n/n! are 0. (These are the limits of the terms, not those of the ratio, although the
limit of the ratio in part b happens to be 0 too.) This example illustrates that n1,000 increases with n so much slower than 3n does
that the series of n1,000/3n converges, and that 3n increases so much slower than n! does that the series of 3n/n! converges.
The power on n used is 1,000. Clearly it can be any positive integer, no matter how large, and the series in part a will still be
convergent. The base of the exponential in n is 3. Clearly it can be any real number (in Part a it must be greater than 1), and
the two series will still be convergent. Thus we see that when n approaches infinity, the power on n approaches infinity
extremely slower than the exponential in n, which in turn approaches infinity extremely slower than the factorial of n.
Relative Speeds Of Increase
We say that quantity Q1 “dominates” quantity Q2 for large n if one of the following is true:
Determine the convergence or divergence of the following series.
When n gets larger and larger, n!/nn gets smaller and
smaller towards 0 so fast that we expect to find out that the series
converges. Observe that nn is a combination of the power and the exponential.
a. Let an = (3n)!/(n!)3. Then:
b. Let an = n!/nn. Then:
Thus, by the RaT,
as 0 < 1/e < 1, the given series converges.
Establish the convergence or divergence of the series:
Observe that in the above solution we didn't seem to employ
any test at all. However we actually did use the SCT: we compare
the given series in part a to the series of (16/27)n and that in part b to the series of (81/64)n.
a. For many series, you should be able to ascertain
convergence or divergence of each at a glance before attempting to use a
test to discover its convergence or divergence.
b. The SCT and the LCT involve two series and compare
their terms. The RooT and the RaT involve only one series. The RooT
examines the limit of the nth root of the general term an of the series. The RaT investigates the limit of the ratio an+1/an of the
terms of the series.
Problems & Solutions
1. Test the convergence of the following series.
2. Determine the convergence or divergence of this series:
This reminds us that before we try to apply a test, we should be alert for series whose terms don't approach 0.
3. Establish the convergence or divergence of the series:
4. Find out whether the following series converges or diverges.
5. Discover the behavior (convergence or divergence) of each series:
a. Let an = 8n+1n2/32n–1. Then:
6. Test the following series for convergence.
Let an = 1,000n(n2 + 10)/n!. Then:
n! “dominates” both 1,000n and n2 + 10.
7. Use the RaT to determine whether this series converges or diverges:
and n!/1,000n approaches infinity as n approaches infinity, the series must diverge to infinity.
Let an = (2n)!/1,000nn!. Then:
Hence, the given series diverges to infinity, by the RaT.
8. Consider the series:
Does the given series converge or diverge? Why?
a. Let an = 22n(n!)2/(2n)!. Then:
The RaT can't establish the convergence or divergence of the given series, because the limit of the ratio is 1.
The given series
diverges to infinity, because the last expression on the right-hand side
approaches 1, and thus the
sequence of the terms of the series doesn't approach 0.