Series are sequences added up. Sequences are a list of numbers in order. For examepl, the sequence `color(black)(a_n=a_1+(n-1)d)` is the equation for a arithmetic sequence. If `color(black)(a_1=3, d=2)`, then the sequence would be {3, 5, 7, 9...}. If you added up all those numbers in the sequence, then you would have a series.
Notation for the series of the arithmetic sequence would be `color(black)(sum_(n=1)^oo 3+2(n-1))`
This series would be divergent because it would go to infinity. If a series equals a certain number, it is convergent.
Using these tests will tell you whether or not a series is convergent or divergent.
Test | Series | Condition(s) of Convergence | Condition(s) of Divergence | Additional Comments |
---|---|---|---|---|
nth-Term | `color(black)(sum_(n=1)^ooa_n)` | `color(black)()` | `color(black)(lim_(ntooo)a_n!=0)` | Cannot show convergence, only divergence. |
Geometric Series | `color(black)(sum_(n=0)^ooar^n)` | `color(black)(0 < abs(r) < 1)` | `color(black)(abs(r) >= 1)` | If convergent then `color(black)(Sum =a/(1-r)` |
Telescoping Series | `color(black)(sum_(n=1)^oo(b_n-b_(n+1)))` | `color(black)(lim_(ntooo)b_n=L)` | `color(black)()` | `color(black)(Sum = b_1-L)` |
p-Series | `color(black)(sum_(n=1)^oo1/(n^p))` | `color(black)(p > 1)` | `color(black)(0 < p <= 1)` | `color(black)()` |
Alternating Series | `color(black)(sum_(n=1)^oo(-1)^(n-1)a_n)` | `color(black)(a_n)` is decreasing, `color(black)(lim_(ntooo)a_n=0)` and `color(black)(a_n > 0)` for all n | `color(black)()` | Remainder `color(black)(abs(R_N) <= a_(N+1))` |
Integral Test | `color(black)(sum_(n=1)^ooa_n, a_n=f(n)>=0)` | `color(black)(int_1^oof(x)dx)` converges | `color(black)(int_1^oof(x)dx)` diverges | Can only be used if f(x) is continuous, positive and decreasing. Remainder: `color(black)(0 < R_N < int_N^oof(x)dx)` |
Root | `color(black)(sum_(n=1)^ooa_n)` | `color(black)(lim_(ntooo)root(n)(abs(a_n))<1)` | `color(black)(lim_(ntooo)root(n)(abs(a_n))>1)` or `color(black)(=oo)` | Test is inconclusive when `color(black)(root(n)(abs(a_n))=1)` |
Ratio | `color(black)(sum_(n=1)^ooa_n)` | `color(black)(lim_(ntooo)abs((a_(n+1))/a_n) < 1)` | `color(black)(lim_(ntooo)abs((a_(n+1))/a_n) > 1)` or `color(black)(=oo)` | Test is inconclusive when `color(black)(lim_(ntooo)abs((a_(n+1))/a_n) = 1)` |
Direct Comparison `color(black)((a_n),b_n>0)` | `color(black)(sum_(n=1)^ooa_n)` | `color(black)(0 < a_n <= b_n)` and `color(black)(sum_(n=1)^oob_n)` converges | `color(black)(0 < b_n <= a_n)` and `color(black)(sum_(n=1)^oob_n)` diverges | |
Limit Comparison `color(black)((a_n),b_n>0)` | `color(black)(sum_(n=1)^ooa_n)` | `color(black)(lim_(ntooo)a_n/b_n=L>0)` and `color(black)(sum_(n=1)^oob_n)` converges | `color(black)(lim_(ntooo)a_n/b_n=L>0)` and `color(black)(sum_(n=1)^oob_n)` diverges |
When you find an alternating series that converges, it only conditionally converges until you determine that it absolutely converges. Absolute convergence means that the series will also converge even after you take the absolute value of the series. If the series does not absolutely converge, it can still conditonally converge or diverge.
For example, `color(black)(sum_(n=1)^oo(-1)^n 1/n)` conditionally converges due to the alternating series test because the series 1/n exists for all n > 1, it is decreasing, and it approaches 0. It is not absolutely converge because `color(black)(sum_(n=1)^ooabs((-1)^n 1/n)=sum_(n=1)^oo 1/n)` diverges due to p-series because it is `color(black)(1/(n^1))` and according to p-series it diverges.
If you have an alternating series and you find the Nth partial sum of a converging alternating series, then `color(black)(abs(S_oo - S_N) <= abs(a_(N+1)))`
Question 1: Is `color(black)(sum_(n=1)^oo((n+2)(2n-5)^4)/(13n^6))` convergent or divergent?
Question 2: Is `color(black)(sum_(n=1)^oo(n^2-1)/(n+1))` convergent or divergent?
Question 3: Can you use the integral test on `color(black)(sum_(n=1)^oo1/(n^2-9))`?
Question 4: Is `color(black)(sum_(n=1)^oo1/(n^4))` convergent or divergent?
Question 5: Is `color(black)(sum_(n=1)^oon/(n-2))` convergent or divergent?
Question 6: Is `color(black)(sum_(n=1)^oo(1/(3^n+8)))` convergent or divergent?
Question 7: Is `color(black)(sum_(n=1)^oo(-1)^n(1/n^4))` absolutely convergent, conditionally convergent or divergent?
Question 8: Is `color(black)(sum_(n=1)^oo(n+4)/(n!))` convergent or divergent?
Question 9: Is `color(black)(sum_(n=1)^oo((3n+1)/(4-2n))^(2n))` convergent or divergent?
Question 10: Is `color(black)(sum_(n=1)^oo7(2/3)^n)` convergent or divergent?
Click on the up carrot to get the answer.
Answer 1: Inconclusive (nth-Term)
Answer 2: Divergent (nth-Term)
Answer 3: No; the series is undefined at n=3 (Integral)
Answer 4: Convergent (p-series)
Answer 5: Divergent (Direct comparison test; compare to 1/n, 1/n is smaller than the series and 1/n diverges, so the series diverges)
Answer 6: Convergent (Limit comparison test; compare to `color(black)(1/(3^n))`, `color(black)(1/(3^n))` converges due to geometric series test, and `color(black)((1/(3^n))/(1/(3^n+8))=1)` which is positive real number.
Answer 7: Absolutely Convergent (Alternating series test)
Answer 8: Convergent (Use ratio test and you will get `color(black)(lim_(ntooo)abs((a_(n+1))/a_n) = 0)`)
Answer 9: Divergent (Root test)
Answer 10: Convergent (Geometric test)
Power series is an infinite series that is an exact representation of an elementary function.
Power series take the form of `color(black)(sum_(n=1)^ooa_nx^n)` or `color(black)(sum_(n=1)^ooa_n(x-c)^n)`
Power series can either converge at one point, for all x, or only over an interval.
If you are given the power series `color(black)(sum_(n=1)^ooa_n(x-c)^n)` then this means that the series either converges only at x=c, or for all x, or on an interval about c with radius R so that if `color(black)(abs(x-c) < R)` then the series converges absolutely. If `color(black)(abs(x-c) > R)` then the series diverges.
If a function `color(black)(f(x) = sum_(n=1)^ooa_n(x-c)^n)` has a radius of convergence of R > 0, then on the interval (c-R, c+R) f is differentiable and also continuous. In that case, `color(black)(f'(x) = sum_(n=1)^oona_n(x-c)^(n-1))` or `color(black)(intf(x)dx = c + sum_(n=1)^oo(a_n(x-c)^(n+1))/(n+1))`
Something to note is that when differentiating or integrating, the radius of convergence of the series obtained by doing such operations is the same as that of the original power series. However, the interval of convergence may be different.
If `color(black)(f(x)=sum_(n=0)^ooa_nx^n)` and `color(black)(f(x)=sum_(n=0)^oob_nx^n)` then:
1. `color(black)(f(kx)=sum_(n=0)^ooa_nk^nx^n)`
2. `color(black)(f(x^N)=sum_(n=0)^ooa_nx^(Nn))`
3. `color(black)(f(x)+-g(x)=sum_(n=0)^oo(a_n+-b_n)x^n)`
You can find the radius of convergence through the use of the ratio test, because you will end up with the form `color(black)(abs(x-c)` > 1. After you find your interval of convergence, you must test the end points to see if the series is convergent at the end points.
The sum of a geometric series is written as `color(black)(Sum =a/(1-r)`. If you are given something in the form of `color(black)(a/(1-r)` then you can rewrite it in a the form of a power series as `color(black)(sum_(n=0)^ooar^n)`.
For example, if you are given the equation `color(black)(3/(2+x^2)` which is equal to `color(black)(3/(2(1-(-1/2x^2)))` which equals `color(black)((3/2)/((1-(-1/2x^2))))`. When you put this in the form of a power series you will get `color(black)(sum_(n=0)^oo(3/2)(-1)^n(1/2x)^(2n))`.
Taylor series is a power series that represents a function centered around the x=c. Taylor series are generally in the form of
`color(black)(sum_(n=0)^oo f^(n)(c) (x-c)^n/(n!))` `color(black)(=f(c)+(f'(c)(x-c))/(1!)+(f''(c)(x-c)^2)/(2!)+...)`
You can see that the Taylor series representation becomes more accurate as you have more terms. If you click on edit graph in the graph below and slide the a= slider, you can see how the representation becomes more accurate.
If `color(black)(P_n(x))` is the nth degree Taylor polynomial of f(x) with a center at x=c then the error will be
`color(black)(abs(f(x)-P_n(x)) <= M/((n+1)!)(abs(x-c))^(n+1))`
Where `color(black)(abs(f^(n+1)(a))<=M)` or maximum for all a between x and c.
Write the Taylor series representation for `color(black)(f(x)=e^(-x))`
Click on the up carrot to get the answer.
Answer: `color(black)(e^(-x)=sum_(n=0)^oo(-1)^(n)(e^4(x+4)^n)/((n)!))`
MacLaurin Series are power series that represents a function centered around x=0, so they are in the form of
`color(black)(sum_(n=0)^oo f^(n)(0) (x)^n/(n!))` `color(black)(=f(0)+(f'(0)(x))/(1!)+(f''(0)(x)^2)/(2!)+...)`
Here are some MacLaurin series representation of functions (centered around 0) that you will need to know for the AP test.
`color(black)(e^x=sum_(n=0)^oo(x^n)/(n!))`
`color(black)(ln(x+1)=sum_(n=1)^oo(-1)^(n-1)(x^n)/(n))`
`color(black)(sin(x)=sum_(n=1)^oo(-1)^(n-1)(x^(2n-1))/((2n-1)!))`
`color(black)(cos(x)=sum_(n=0)^oo(-1)^(n)(x^(2n))/((2n)!))`
Write the MacLaurin series representation for `color(black)(f(x)=cos(4x))`
Click on the up carrot to get the answer.
Answer: `color(black)(cos(4x)=sum_(n=0)^oo(-1)^(n)(16^nx^(2n))/((2n)!))`