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SECTION 9.2: ARITHMETIC SEQUENCES and PARTIAL SUMS
PART A: WHAT IS AN ARITHMETIC SEQUENCE?
The following appears to be an example of an arithmetic (stress on the โmeโ) sequence: a 1 = 2 a 2 = 5 a 3 = 8 a 4 = 11 ๏ We begin with 2. After that, we successively add 3 to obtain the other terms of the sequence. An arithmetic sequence is determined by:
- Its initial term Here, it is a 1 , although, in other examples, it could be a 0 or something else. Here, a 1 = 2.
- Its common difference This is denoted by d. It is the number that is always added to a previous term to obtain the following term. Here, d = 3. Observe that: d = a 2 โ a 1 = a 3 โ a 2 = โฆ = ak + 1 โ ak k โZ +
( ) =^ โฆ
The following information completely determines our sequence: The sequence is arithmetic. (Initial term) a 1 = 2 (Common difference) d = 3
In general, a recursive definition for an arithmetic sequence that begins with a 1 may be given by: a 1 given
ak + 1 = ak + d ( k โฅ 1 ; "k is an integer" is implied)
Example The arithmetic sequence 25, 20, 15, 10, โฆ can be described by: a 1 = 25 d = โ 5
PART C : FORMULA FOR THE n th PARTIAL SUM OF AN ARITHMETIC SEQUENCE The n th partial sum of an arithmetic sequence with initial term a 1 and common difference d is given by: Sn = n a 1 + an 2
Think: The (cumulative) sum of the first n terms of an arithmetic sequence is given by the number of terms involved times the average of the first and last terms. Example Find the 100 th partial sum of the arithmetic sequence: 2, 5, 8, 11, โฆ Solution We found in the previous Example that: a 100 = 299 Sn = n a 1 + an 2
S 100 = ( 100 )
i.e., 2 + 5 + 8 + ... + 299 = 15 , 050 This is much easier than doing things brute force on your calculator! Read the Historical Note on p.628 in Larson for the story of how Gauss quickly computed the sum of the first 100 positive integers, k k = 1 100
โ =^1 +^2 +^3 +^ ...^ +^100. Use our
formula to confirm his result. Gaussโs trick is actually used in the proof of our formula; see p.694 in Larson. We will touch on a related question in Section 9.4.
SECTION 9.3: GEOMETRIC SEQUENCES, PARTIAL SUMS, and
SERIES
PART A: WHAT IS A GEOMETRIC SEQUENCE?
The following appears to be an example of a geometric sequence: a 1 = 2 a 2 = 6 a 3 = 18 a 4 = 54 ๏ We begin with 2. After that, we successively multiply by 3 to obtain the other terms of the sequence. Recall that, for an arithmetic sequence, we successively add. A geometric sequence is determined by:
- Its initial term Here, it is a 1 , although, in other examples, it could be a 0 or something else. Here, a 1 = 2.
- Its common ratio This is denoted by r. It is the number that we always multiply the previous term by to obtain the following term. Here, r = 3. Observe that: r = a 2 a 1
a 3 a 2
ak + 1 ak k โZ
( ) =^ โฆ
The following information completely determines our sequence: The sequence is geometric. (Initial term) a 1 = 2 (Common ratio) r = 3
PART B : FORMULA FOR THE GENERAL n th TERM OF A GEOMETRIC SEQUENCE Letโs begin with a 1 and keep multiplying by r until we obtain an expression for an , where n โZ
. a 1 = a 1 a 2 = a 1 โ
r a 3 = a 1 โ
r 2 a 4 = a 1 โ
r 3 ๏ an = a 1 โ
r nโ 1 The general n th term of a geometric sequence with initial term a 1 and common ratio r is given by: an = a 1 โ
r nโ 1 Think: As with arithmetic sequences, we take n โ 1 steps to get from a 1 to an. Note: Observe that the expression for an is exponential in n. This reflects the fact that geometric sequences often arise from exponential models, for example those involving compound interest or population growth.
Example Find the 6 th term of the geometric sequence: 2, โ 1 ,
(Assume that 2 is the โfirst term.โ) Solution Here, a 1 = 2 and r = โ
an = a 1 โ
r nโ 1
a 6 = ( 2 ) โ
6 โ 1
5
Observe that, as n โ โ , the terms of this sequence approach 0. Assume a 1 โ 0. Then, a 1 โ
r nโ 1
( โ^0 as^ n^ โ^ โ) โ^ ( โ^1 <^ r^ <^1 )
i.e., r < 1
Example Find the 6 th partial sum of the geometric sequence 2, โ 1 ,
Solution Recall that a 1 = 2 and r = โ
for this sequence. We found in the previous Example that: a 6 = โ
We will use our formula to evaluate: S 6 = 2 โ 1 +
Using our formula directly: Sn = a 1 โ a 1 r n 1 โ r or a 1 1 โ r n 1 โ r
If we use the second version on the right โฆ Sn = a 1 1 โ r n 1 โ r
S 6 = 2
6 1 โ โ
21 63 32 64
1 (^31)
We can also use the first version and the โfirst in โ first outโ idea: S 6 = 2 โ 1 +
โFirst outโ is: a 7 =
Sn = a 1 โ a 1 r n 1 โ r S 6 =
21 63 16 32
1 (^31) =
Example The geometric series 2 + 6 + 18 + 54 + ... has no sum, because: lim nโโ Sn = โ Example The geometric series 1 โ 1 + 1 โ 1 + ... has no sum, because the partial sums do not approach a single real number. Observe: 1 S 1 = 1
S 2 = 0
S 3 = 1
S 4 = 0
An infinite geometric series converges โ ( โ 1 < r < 1 )
i.e., r < 1
Take another look at the Examples of this Part. It is true that an infinite geometric series converges โ Its terms approach 0. Warning: However, this cannot be said about series in general. For example, the famous harmonic series
k = 1 k โ
โ =^1 +^
- โฆ does not converge, even though the terms of the series approach 0. In order for a series to converge, it is necessary but not sufficient for the terms to approach 0. No infinite arithmetic sequence (such as 2 + 5 + 8 + 11 + โฆ ) can have a sum, unless you include 0 + 0 + 0 + ... as an arithmetic sequence.
The sum of a convergent infinite geometric series with initial term a 1 and common ratio r , where โ 1 < r < 1 i.e., r < 1
, is given by: S = a 1 1 โ r Technical Note: This comes from our partial sum formula Sn = a 1 โ a 1 r n 1 โ r and the fact that a 1 r n
( โ^0 as^ n^ โ^0 ) if^ โ^1 <^ r^ <^1
i.e., r < 1
Example Write 0. 81 as a nice (simplified) fraction of the form integer integer
Recall how the repeating bar works: 0. 81 = 0.81818181... Note: In Arithmetic, you learned how to use long division to express a โniceโ fraction as a repeated decimal; remember that rational numbers can always be expressed as either a terminating or a โnicelyโ repeating decimal. Now, after all this time, you will learn how to do the reverse! Solution
- 81 can be written as: 0.81 + 0.0081 + 0.000081 + ... Observe that this is a geometric series with initial term a 1 = 0.81 and common ratio r =
= 0.01; because r < 1 , the series converges. The sum of the series is given by: S = a 1 1 โ r
