Chapter 6 series Flashcards
Definition 6.11:
Partial sums
SUMMABLE
Given a sequence (a_n) of real numbers we consider the associated sequence (s_n) if partial sums
Where
s_n = a₁ + a₂ + … + a_n
We say that the sequence (a_n) is SUMMABLE if the sequence (s_n) of partial sums converges.
Writing the infinite series from n=1 as the limit of s_n as n tends to infinity
Limit as s_n tends to infinity = a₁ + a₂ + …
Geometric series
Sum from n=0 to infinity of a•r^n which converges for mod r less than 1.
BEGINS AT N=0 rather than n=1
Converges to a/ (1-r) WHEN STARTS AT n=0!!!
Because…
Partial sums are
s_n = a(1-r^n)/ (1-r)
Taking its limit
Proposition 6.1.3
“(The divergence test)”
Suppose the series
Sum from n=1 to Infinity
Of a_n
Converges
Then a_n tends to 0 as n tends to infinity.
Converse not true: the harmonic series doesn’t converge but 1/n does tend to 0
Example: show that the series
Sum from n=1 to infinity
Of
n/ (n+2)
Doesn’t converge
The limit as n tends to infinity of
n/(n+2) =
Limit as n tends to infinity of 1/(1+ 2/n) =1 not equal to 0. Hence by proposition 6.1.3 the series doesn’t converge.
Definition 6.1.5 absolute convergence
A series Sum from n=1 to infinity Of a_n converges absolutely when the series Sum from n =1 to infinity of |a_n| converges
Proposition 6.1.5
Absolute convergence implies
any series that converges absolutely
also converges
Proof: given that converges absolutely we use Cauchy and show there exists n and m for this property. Hence showing that |s_n -s_m| also less than epsilon for showing convergence of non absolute series
Thus (s_n) us Cauchy and hence convergent.
Types of convergence tests
Comparison test
Ratio test
Root test
“Divergence test”
Theorem 6.2.1 comparison test
Let (a_n) and (b_n) be sequences of real numbers where a_n bigger it equal to 0 and b_n bigger or equal to 0.
Suppose that (b_n) is summable and a_n is less than or equal to b_n for all sufficiently large n. Then (a_n) is summable.
Theorem 6.2.4 ratio test
Consider a series of non zero terms:
Sum from n=1 to infinity of a_n
And suppose that
Limit as n tends to infinity of | (a_n +1)/(a_n)| =r
If r is less than 1 series CONVERGES ABSOLUTELY
if r is bigger than 1 the series does not converge.
r=1 tells us nothing.
Theorem 6.2.6 root test
Let the sum of n=1 to infinity
Of a_n be a series
and set c = limit as n tends to infinity
Of
Sup{ |a_n|^(1/n), |a_{n+1}|^{1/(n+1)}, |a_{n+2}|^(1/(n+2)),…}
If c is less than 1 then the series CONVERGES ABSOLUTELY
If c is bigger than 1 then the series does not converge.
Example show that the series converges
Sum from n=1 to infinity of n3^ {-n}
Observe that for all n in the naturals n is less than or equal to 2^n.
Hence n3^-n is less than or equal to
(2/3)^n
and the series
Sum from n=1 to infinity of (2/3)^n is a geometric series with common ratio less than 1 hence converges and by the comparison og series converges
Example 6.2.5 let x be in R.
Investigate convergence of the series
Sum from n=1 to infinity of x^n / n
Using the ratio test
Let a_n = x^n /n. Then
a_{n+1} / a_n
= [(x^ {n+1}) / (n+1)] /[ x^n /n]
= nx / (n+1) = (x)/( 1 + 1/n)
So the limit as n tends to infinity of
| a_n+1 / a_n| = |x|
Thus by the ratio test our series CONVERGES ABSOLUTELY if |x|<1 and doesn’t converge if bigger than 1.
Tells us nothing at 1 or -1: converges but not absolutely at -1 and doesn’t at 1
For any bounded sequence of numbers the sequence of the supremum of these …
For any bounded sequence of numbers (b_n) where b_n is bigger than or equal to 0.
The sequence (c_n) where
c_n = sup { b_n, b_n+1, b_n+2,…}
Is BOUNDED and monotone decreasing, so sequence c_n converges.
If (b_n)wasn’t bounded then limit of c_n Is infinity??
