Quiz 6B for MATH 105 SECTION 205
March 27, 2015
Given Name
Family Name
Student Number
∞
X
nan − 2n + 1
1. (10 points) Assume that the series
− ln a1 +
∞
X
n=1
∞
X
ln
n=1
ln
an
an+1
converges, where an > 0 for all n ≥ 1. Is the series
n+1
n=1
convergent or divergent. If it’s convergent, evaluate the sum of the series − ln a1 +
an
.
an+1
Answer.
For any m ≥ 1, let Sm = − ln a1 +
m
X
an
, then
an+1
ln
n=1
Sm = − ln a1 +
= − ln a1 +
m
X
n=1
m
X
ln
an
an+1
[ln an − ln an+1 ]
n=1
− ln a1
=
+ ln a1 − ln a2
+ ln a2 − ln a3
+ ln a3 − ln a4
..
.
+ ln am−1 − ln am
+ ln am − ln am+1
= − ln am+1
So we should compute lim ln am+1 . Since that the series
m→∞
we must have lim
n→∞
nan − 2n + 1
= 0. Hence we get
n+1
lim
n→∞
Since lim
n→∞
∞
X
nan − 2n + 1
n=1
n+1
n(an − 2)
= 0.
n+1
n+1
= 1, then we have
n
lim
n→∞
n + 1 n(an − 2)
·
= 0.
n
n+1
That is, we have
lim an = 2.
n→∞
Hence we get
lim Sm = − lim ln am+1 = − ln 2.
m→∞
m→∞
converges, by the divergence test,
In summary, we know that the series − ln a1 +
∞
X
ln
n=1
− ln a1 +
∞
X
an
an+1
ln
n=1
an
an+1
is convergent and
= − ln 2.
Remark 1: For example, let
an = 2 −
1
1
+ ,
n n2
for all n ≥ 1,
Then we have
nan − 2n + 1
n
So we get
n
1
(an − 2) +
n+1
n+1 n
1
1
1
=
2− + 2 −2 +
n+1
n n
n+1
n
1
n
1
1
= −
· +
· 2+
n+1 n n+1 n
n+1
1
=
.
n(n + 1)
=
∞
X
nan − 2n + 1
n=1
n
=
∞
X
n=1
1
= 1.
n(n + 1)
Your Score:
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