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Unformatted text preview: nclude that
np lim nÝ jn1 1 log p.
j log j Some Exercises on The Comparison Test
Test each of the following series for convergence.
1
n 3/2 n 1. Solution: We define
an 1
n 3/2 n and
1
n 3/2
b n is a convergent pseries and since
bn for each n. Since the series 1
a n lim n 3/2 n lim n 3/2
lim
n Ý bn
nÝ
n Ý n 3/2 n
1
n 3/2
1
nÝ
lim
1
1 1n
we see that a n is convergent. Note that we could have achieved this solution more rapidly by observing that a n b n for each n.
However, taking the limit was worth while because it applies just as well to Exercise 2.
2. 1
n 3/2 n
We define
an 1
n 3/2 n and
1
n 3/2
for each n. Since
b n is a convergent pseries and since
lim a n 1
n Ý bn
a n is convergent.
we deduce that
bn 3. n4 n
n2 2 Solution: In order to see how to proceed observe that, for large n, the expression
n4 n
n2 2 behaves like
n 1
n
n4
because n 2 2 is much smaller than n 4 . Now we begin. 324 We define
an n4 n
n2 2 and
for each n. Since we deduce that bn 1
n
b n is a divergent pseries and since
n
4
n2 2
n
lim a n n Ý
lim
nÝ
lim
1
n Ý bn
n n4 n4
1
n2 2 a n is divergent. n log n 4. n2 2 n5 Solution: The key to this exercise is the fact that, although n log n is larger than n, it is not much
larger because of the fact that
log n
0
n lim nÝ whenever 0. The denominator of the fraction
n log n
n2 2 n5
behaves like n 5/2 . We can therefore consider that n log n
n2 2 n5
behaves like
log n
n log n
3/2
n
n 5/2 log n
n 1/4 1
n 5/4 . With this in mind, we define
n log n an n5 n2 2 and
1
n 5/4
b n is a convergent pseries and that
log n
lim a n n Ý 1/4 0
lim
n Ý bn
n
bn for each n. We observe that a n is convergent. and so
5. 1 n 1 log n /n Solution: We recall that if a and b are any positive numbers then
a b exp b log a .
We begin by observing that if n is any positive integer then
log n
log n
n 1 log n /n n n log n /n n exp
n
Since 325 n exp log n
n 2 log n
n lim exp nÝ 2 log n
n exp n Ý
lim 2 exp 0 1 we conclude that
1 1 log n /n n
behaves like 1/n for large n . We define 1 an n 1 log n /n and
for each n. Since bn 1
n
b n is a divergent pseries and since
1
lim a n n Ý
lim
n Ý bn
log n
exp 1 2 n we conclude that a n is divergent. 1 6.
n 1 log n 2 /n Hint: This exercise is similar to Exericse 5 because of the identity
n 1 log n 2 /n 3 log n
n n exp and because
log n
n lim exp nÝ 1 Thus
n 3 log n
n exp n Ý
lim 3 exp 0 1. is divergent. 1 log n 2 /n 1 7.
n 1 log n log log n /n Solution: For each n we see that
n 1 log n log log n n exp /n log n log n
n log log n Now
log n log log n exp log log n 2 log log n
log n exp 2 and since
lim nÝ log log n
log n 2 0 we conclude that if n is sufficiently large then
log n log log n exp log log n
log n exp 1 log n
2
Therefore, for n sufficiently large we have
log n log n
0
n log log n and we conclude that 326 2 log n n 1/2
log n n 1/2
log n
1/2
n
n log n log n lim
nÝ log log n...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
 Fall '08
 STAFF
 Math, Calculus

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