Calculus Of One Real Variable – By Pheng Kim Ving


Return To Contents
Go To Problems & Solutions
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 x^{1/}^{x} As x Approaches Infinity
For real variable x: 
Proof
EOP
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)

Proof
EOP
Note that the proof of the RooT makes use of the SCT.
Example 1.1
Test the convergence of the series:
Note
Solution
EOS
The RooT, which tests the nth root of the general term, is especially useful when the general term is an nth power.
Example 1.2
Determine the convergence/divergence of these series using the RooT:
Note
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.
Solution
EOS
Return To Top Of Page Go To Problems & Solutions
2. The Ratio Test (RaT) 
Consider the geometric series:
The examination of the limit of the ratio a_{n}_{+1}/a_{n} 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)

Proof
The two series
have the same ratio limit of 1, but one converges while the other diverges.
This completes the proof.
EOP
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.
Example 2.1
Test each of the following series for convergence:
Note
Solution
a. Let a_{n} = n^{1,000}/3^{n}. Then:
EOS
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.
Contributed
Factors
Remark that n^{1,000}
contributes a factor equivalent to 1 (= 1^{1,000}) in the expression of the last
limit, 3^{n} 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
(n^{1,000}) and an exponential in n (3^{n}). Part b involves an
exponential and a factorial. Since both
series converge, the limits of n^{1,000}/3^{n}
and 3^{n}/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 n^{1,000} increases with n so much slower than 3^{n}
does
that the series of n^{1,000}/3^{n}
converges, and that 3^{n} increases so much
slower than n! does that the series of 3^{n}/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 Q_{1} “dominates” quantity Q_{2} for large n if one of the following is true:
Example 2.2
Determine the convergence or divergence of the following series.
Note
When n gets larger and larger, n!/n^{n} gets smaller and
smaller towards 0 so fast that we expect to find out that the series
converges. Observe that n^{n} is a combination of
the power and the exponential.
Solution
a. Let a_{n} = (3n)!/(n!)^{3}. Then:
b. Let a_{n} = n!/n^{n}. Then:
Thus, by the RaT,
as 0 < 1/e < 1, the given series converges.
EOS
Example 2.3
Establish the convergence or divergence of the series:
Solution
EOS
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}.
Return To Top Of Page Go To Problems & Solutions
3. Remarks 
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 a_{n}
of the series. The RaT investigates the limit of the ratio a_{n}_{+1}/a_{n} of the
terms of the series.
Problems & Solutions 
1. Test the convergence of the following series.
Solution
2. Determine the convergence or divergence of this series:
Solution
Note
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:
Solution
4. Find out whether the following series converges or diverges.
Solution
5. Discover the behavior (convergence or divergence) of each series:
Solution
a. Let a_{n} = 8^{n}^{+1}n^{2}/3^{2}^{n}^{–1}. Then:
6. Test the following series for convergence.
Solution
Let a_{n} = 1,000^{n}(n^{2} + 10)/n!. Then:
Note
n! “dominates” both 1,000^{n} and n^{2} + 10.
7. Use the RaT to determine whether this series converges or diverges:
Note
Since:
and n!/1,000^{n} approaches infinity as n approaches infinity, the series must diverge to infinity.
Solution
Let a_{n} = (2n)!/1,000^{n}n!. Then:
Hence, the given series diverges to infinity, by the RaT.
8. Consider the series:
Does the given series converge or diverge? Why?
Solution
a. Let a_{n} = 2^{2}^{n}(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 righthand side
approaches 1, and thus the
sequence of the terms of the series
doesn't approach 0.
Return To Top Of Page Return To Contents