Exercise 8.4.3
Since | r | = 3 / 5 < 1, the series converges to
Exercise 8.4.2
Find the sum of the first ten terms:
Find the sum of the first ten terms:
Exercise 8.4.1
Exercise 8.3.5
There are 12 – 2 + 1 = 11 terms.
The first term is 2 ( 2 ) – 1 = 3.
The last term is 2 ( 12 ) – 1 = 23.
The sum equals the sum of the first and last term multiplied by half the number of terms.
Sum = ( 3 + 23 )( 11 / 2 ) = ( 13 )( 11 ) = 143
Exercise 8.3.4
To find the sum of the first 70 terms of the arithmetic sequence –11, -6, -1, 4, . . ., we need to find out what the last term equals. The first terms is –11 and the common difference is 5, so the 70th term equals –11 plus 69 times 5, or 334. The sum of the arithmetic sequence equals the sum of the first and last terms times half the number of terms, so the sum equals ( -11 + 334 )( 35 ) = 11,305.
Exercise 8.3.3
We know that, in general, the formula for the nth term of an arithmetic sequence with first term c1 and common difference d is cn = c1 + ( n – 1 ) d.
So c3 = c1 + 2 d = 8
and c9 = c1 + 8 d = 56
This gives us two equations in the two unknowns c1 and d.
Subtracting the first equation from the second gives 6 d = 48, so the common difference is 8. Substituting d = 8 into the first equation and solving tells us that the first term c1 equals – 8.
Exercise 8.3.2
c27 = c1 + 26 d = 2 + 26 ( - 3 ) = - 76.
Exercise 8.3.1
For the sequence 2, 5, 8, 1, . . .
c1 = 2
cn+1 = cn + 3
For the sequence –1, 3, 8, 13, . . .
c1 = - 2
cn+1 = cn + 5
Exercise 8.2.5
The particular letter used for the index of a sum doesn't matter. So the letter q used to index the second sum may be replaced with the letter p, making this problem just like problem 8.2.4.
The answer is 
.
Exercise 8.2.4
Exercise 8.2.3
( 32 – 2 ) + ( 42 – 2 ) + ( 52 – 2 ) + ( 62 – 2 ) + ( 72 – 2 ) = 7 + 14 + 23 + 34 + 47 = 125
Exercise 8.2.2
( 1 – 1 )2 + ( 2 – 1 )2 + ( 3 – 1 )2 + ( 4 – 1 )2 = 0 + 1 + 4 + 9 = 14
Exercise 8.2.1
Notice that the numerators are just the numbers 1, 2, 3, . . . and the denominators are one more than the numerators.
Exercise 8.1.5
All the numerators equal 1 and the denominators are powers of 2, beginning with 20 as the denominator of the first term. So the nth term is cn = 1 / 2n.
Look at the sequence of partial sums:
s1 = 1 = 2 - 1
s2 = 1 + 1 / 2 = 3 / 2 = 2 – 1 / 2
s3 = 1 + 1 / 2 + 1 / 4 = 7 / 4 = 2 – 1 / 4
s4 = 1 + 1 / 2 + 1 / 4 + 1 / 8 = 15 / 8 = 2 – 1 / 8
So the apparant pattern is sn = 2 – 1 / 2n-1.
To prove that 2 is the limit of the sequence of partial sums, let ( a, b ) denote an interval containing the number 2. If we can show that the interval must contain a tail of the sequence { 2 – 1 / 2n-1 }, then 2 must be the limit of the sequence. Since the only condition we have place on the interval ( a, b ) is that it contain 2, anything we conclude must be true for any interval containing 2, no matter how small.
The key lies in the fact that the number 1 / 2n-1 may be made as small as we please just by picking n large enough. Thus, we can pick n large enough so that 1 / 2n-1 < 2 – a. It would follow that a < 2 - 1 / 2n-1 < 2.
Thus the nth term of the partial sums, and every term thereafter would lie in the interval ( a, b ). Thus the interval contains a tail of the sequence. So 2 is the limit of the sequence.
Notice that we could not have done this for any number except 2. Suppose we had thought the limit was 2.5, for example. Is it true that, for any interval ( a, b ) containing 2.5, no matter how small, that interval must contain a tail of the sequence? No, it is not true. Because we can make the interval small enough so that it does not contain 2. Then there is no way the interval could contain any term of the sequence, since all the terms of the sequence are smaller than 2.
OK, you say, the limit can't be larger than 2. Maybe it can be smaller than 2, such as 1.5, for example. Is it true that any interval ( a, b ) containing 1.5 must contain a tail of the sequence? No it isn't. Let the interval be small enough that 1.5 < b < 2. Then n can be made large enough that the interval ( b, 2 ) contains every term of the sequence beginning with the nth term onward. So ( a, b ) cannot contain a tail of the sequence in that case. So no number smaller than 2 can be the limit of the sequence.
Exercise 8.1.4
We want to show that the sequence { 2 + 1 / n } approaches a limit of 2. In order to show this, we must show that any open interval containing 2, no matter how small, must contain a tail of the sequence.
Notice that each term of this sequence is smaller than the prededing term. This is a decreasing sequence.
Let ( a, b ) be any sequence containing 2. We can pick n large enough to make 1 / n as small as we please, so pick n large enough that 1 / n is smaller than b – 2. Then 2 + 1 / n is smaller than b, so the nth term of the sequence and all terms thereafter must lie in the interval ( a, b ). So ( a, b ) contains a tail of the sequence. So 2 is the limit of the sequence.
Exercise 8.1.3
c1 = 1
c2 = 1
c3 = c1 + c2 = 2
c4 = c2 + c3 = 3
c5 = c3 + c4 = 5
Notice that if 1 is subtracted from each term of the sequence, we get the sequence
1, 4, 9, 16, 25, . . . which are all perfect squares. The terms of the sequence
2, 5, 10, 17, 26, . . . are all one more than a perfect square.
The general formula for the nth term is cn = 1 + n2.
Exercise 8.1.1
0, 2, 6, 12, 20