MATH 290 – HOMEWORK 10 – SOLUTIONS
In this homework, we will prove from scratch (that is, without using L’Hopital’s Rule)
that lim
n
1
/n
= 1.
1. Prove that lim
s
2
n

1
= 0. Do this using the definition of convergence, and by invoking
the Archimedean Principle of the real numbers explicitly where it is required.
Solution:
We need to show that given
>
0 there is an
N
∈
N
such that if
n
≥
N
then
s
2
n

1

0
<
.
Let
>
0.
Using the Archimedean Principle of the
real numbers, we can choose
N
∈
N
such that
N >
(2
/
2
) + 1.
If
n
≥
N
,
then
n
≥
N >
(2
/
2
) + 1, so that
n >
(2
/
2
) + 1.
From this it follows that
n

1
>
2
/
2
, that 1
/
(
n

1)
<
2
/
2, that 2
/
(
n

1)
<
2
, and finally that
s
2
n

1
=
s
2
n

1

0
<
as required.
2. Prove the following by induction.
(a) For every
x >
0 and every
n
∈
N
, (1 +
x
)
n
> nx
.
(b) For every
x >
0 and every
n
∈
N
with
n
≥
2, (1 +
x
)
n
>
n
(
n

1)
2
x
2
. (Hint: You may
use the result in part (a) for this proof, and the fact that (
n
+ 1)
n

n
(
n

1) = 2
n
.)
Solution:
(a). If
n
= 1 then the inequality reduces to 1 +
x > x
which is true. Hence the
result holds for
n
= 1.
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 Fall '08
 Alligood,K
 Math, Advanced Math, Binomial Theorem, Mathematical Induction

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