Exercise.

(a) Assuming for the moment that (_{
)
} is
convergent, by solving a quadratic equation or otherwise
make a reasonable conjecture

for a value
that might be the limit of the sequence.

(b) Prove that 1
_{
}
holds
for all , by induction. Your proof must not
use the conjecture from (a) that _{
},
only the value of itself.

(c) Using part (b), or by induction, prove that the sequence is
monotonic nondecreasing.

(d) Which theorem from lectures allows you to deduce from
(b) and (c) that _{
} has a limit?

(e) Apply the continuity of the function
()=
1+
(considered as a function
of positive into the set of positive reals)
to prove that the limit of the sequence (_{
)
}
is , as conjectured in part (a).

Exercise.

For the following series
series
=1
_{
}
, determine if
the series converges and if so find the limit. (Hint: use partial
fractions to express _{
}.)

Exercise.

(a) Let

in

and

.
Prove that
(1+

.
(Use induction on

.)

(b) If
with 0
1
, show that
^{
0
}
as
. (Hint: write
=11+

and use (a).)

(c) Let _{
=
} where 0
1
.
Show that
=1
_{
}
=
(1-)^{2}
(1-(+1)
^{
+
+1
}
) (Use induction on .)

(d) Hence show that the series
=1
^{
}
converges if 0
1
, and find its limit.

(e) Show that
=1
^{
}
diverges if
1
.

Standard results and theorems, including
the squeeze rule,
may be used if quoted correctly.

## Sample solutions to selected exercises

### Exercise 4.3(a)

Note that _{
=1(+1)
}=1
-1+1
so _{
}=11-12+12-13+13-14++
1
-1+1
hence _{
=11-1+1
}
for all . But 1+1
0
as
(being a subsequence of a standard null sequence
1
) and so _{
1-0=1
} by
continuity of - and therefore the
series converges with
=1
_{
=1
}
.

### Exercise 4.4(b)

Since 0
1
there is a real number

such that

=11+
and

.
Now note that 0

^{
=
(1+
)
}
1+
by part (a). Simplifying this we have 0

^{
}
2
2
^{2}
+2
+-
. The sequence (

_{
)
} on the right hand side of this inequality
converges to 0 by the standard null sequence

1
and continuity of addition, subtraction, multiplication, and
of division at (0,

. (Note that

.)
Thus by the squeeze rule applied to 0

^{
0
}
we have

^{
0
}
as

as required.