CK-12-Calculus

8.5. Ratio Test, Root Test, and Summary of Tests http://www.ck12.org

1.∑∞n= 1

( n

n^2 + 1

)n

2.∑∞n= 1

((− 1 )n(lnn)

n

)n

3.∑∞n= 1 ( 53 n+n 2 n)

4. If limn→∞

∣∣

∣ana+n^1

∣∣

∣=α, then limn→∞|√nan|=α.

5. (Hard) Is the series∑∞n= 1

(n (^2) − 1
n^2 + 1
)n
convergent?
Hint:
(n (^2) − 1
n^2 + 1
)n
= n

√√

√√[(n^2 − 1

n^2 + 1

)n 2 ]

and limh→ 0 ( 1 +h)^1 /hexists and equalse.

Summary of Procedures for Determining Convergence

We have seen various test for convergence of∑anin action. To summarize, the key phrase is "recognize the form
ofan".
It is sometimes difficult to choose the best convergence test for a particular series. Not all tests work on any given
series, and even if a test works on a particular series, that test may still involve a lot of work in reaching a convergence
conclusion. With experience, however, we can learn to apply the right test to a given series. At a minimum, we can
learn which tests are easiest to apply, so that we can start with those easier tests if we have no other idea how proceed.
The following is a summary of the various tests and when they might be useful:

TABLE8.2:

Test Form of ofan Comments

1 No/little test geometric, harmonic,

p−test

clear answer

2 Test of divergence limn→∞an 6 = 0 inexpensive test

3 Integral Test corresp. integral in nice

closed form

easy integration

4 Alternating Series ∑(− 1 )nbn (or

∑(−^1 )n−^1 bn)

check conditions onbn

5 (Limit) Comparison need companion known

series

compare

6 Ratio Test∗ recognize good formana+n^1 evaluate ratio

7 Root Test∗ anresemblescnn evaluatenthroot

8 Combination composite of forms combined methods

∗: the inconclusive cases need other tests.
Example 1
∑n(ln^1 n)diverges by the Integral Test since∫x(ln^1 x)dx=ln|ln|x||+Cdiverges.

Example 2∑sin^1 ndiverges by Limit Comparison Test (against the harmonic series) since limn→∞sin (^1) n^1 n=limh→ 0 sinhh=
1 and the harmonic series diverges.
Example 3∑^1 nsin(^1 n)converges with Limit Comparison Test (against∑n^12 ) since limn→∞
(^1) nsin (^1) n
n^12 =limn→∞sin
(^1) n
(^1) n =
limh→ 0 sinhh=1 and the latter series converges.