General Rules
Telescoping and Geometric series are the only types of series that you can estimate sums from. So, you must use these test's properties to estimate these sums |
If the question is asking for absolute convergence or conditional convergence. You will need to use the Ratio Test, Root Test, or the definition of Absolute/Conditional Convergence |
Must show ALL work to receive full credit for questions. Study the process to solve the problems, don't just guess through the review |
Tests
Test for Divergence(TFD) |
Inconclusive |
You absolutely cannot determine if a series is convergent from this test. |
| |
Diverges |
If limit of series ≠0 or ∞ |
Integral Test |
Converges |
If integral of series <∞ |
| |
Diverges |
If integral of series =∞ |
Ratio Test |
Converges/Converges Absolutely |
If limit of 0≤|(a k+1
) /(a k
)|<1 |
| |
Diverges |
If limit >1 |
| |
Inconclusive |
If limit =1 |
Root Test |
Converges/Converges Absolutely |
If 0≤limit of the k th root of |a k
|<1 |
| |
Diverges |
If limit of the k th root of |a k
|>1 |
| |
Inconclusive |
If limit of k th root of |a k
|=1 |
Direct Comparison Test(CT) |
Converges |
If ∑b k
converges AND b k
is the larger of the two functions |
| |
Diverges |
If ∑b k
diverges AND b k
is smaller of two functions |
Limit Comparison Test(LCT) |
Converges |
If b k
converges AND limit of 0<(a k
)/(b k
)<∞ |
| |
Diverges |
If b k
AND limit of 0<(a k
)/(b k
)<∞ |
Alternating Series Test(AST) |
Converges |
If all 3 conditions for AST are met |
| |
Diverges |
If limit condition fails, ∑a k
is immediately divergent by TFD |
|
|
Properties of Special Series
Geometric Series |
Converges |
If Absolute Value of r<1. Converges at S=(first term)/(1-r) |
| |
Diverges |
If Absolute Value of r≥1 |
P-Series |
Converges |
If p>1 |
| |
Diverges |
If p≤1 |
Telescoping |
Converges |
Value that the limit of the remaining terms approach |
| |
Diverges |
Almost never. On the test it will converge |
Definition of Convergence |
Absolute Convergence |
If and only if ∑|a k
| is convergent |
| |
Conditional Convergence |
If and only if ∑a k
is convergent, but ∑|a k
| is divergent |
When to Use Tests
Properties |
If you can identify the series as a geometric, p, or telescoping series, then use their respective properties. If the given series looks close to one of these series see if you can use algebra to rearrange it into one of them |
Test for Divergence(TFD) |
Should at least eyeball this test first to see if the limit of the series does not approach 0. If series does not approach 0, then ∑ak
divergent by TFD |
Comparison Tests(CT and LCT) |
ONLY POSITIVE TERMS! If you can tell if the series has negative terms,((-1) k or sin/cos), do not use this test. If series has is rational and has a root in the denominator, compare with a p-series. |a k
| gives use absolute convergence |
Alternating Series Test(AST) |
Series with (-1)k can be testes with AST |
Integral Test |
ONLY POSITIVE TERMS! If you can look at the function and easily take the integral, it is probably good to use this test |
Ratio Test |
If series contains: k!, or powers and exponentials, almost guaranteed to use ratio test |
Root Test |
If the entire series can be written to the kth power, you can use the root test |
|
|
Integral Test
Conditions:
1. f(x) is positive on its interval
2. f(x) is continuous on its interval
3. f(x) decreasing as x->∞ (derivative is negative)
* Must change a k
to a function in order to take derivative
* Integral starts off from k to ∞, so you must change the integral to k to t with limit as t->∞
* Answer you get is not where the ∑a k
converges |
Alternating Series Test
Conditions:
1. b k
>0
2. b k
≥b k+1
3. limit of b k
=0
* If ∑b k
fits all three conditions, ∑a k
convergent by AST
* If 3 rd condition fails, ∑a k
is divergent by TFD
* If series contains (-m) k, pull (-1) k out and keep m k in b k
|
Integral Remainder
Known nth Value |
Solving for S n
(sum of the series approximation) |
Plug n into the series |
| |
Solving for R n
(approximation of the remainder) |
Solve integral from n to ∞ |
Known Error Bound |
Set the integral of f(x) from n to ∞ less than error bound. Once solved, answer will be given in terms of n<#. Must round up the number since series only use integers and if you rounded down, the value of the integral would be larger than the error bound |
Alternating Series Remainder
Known Error Bound |
Set error bound less than b n+1
. Solve for a #>n, round up n to next highest whole number |
| |
If the inequality is very difficult to solve, the use of a table, shown below, is acceptable. When the middle column, b n+1
, is less than third column, error bound, then that value is you final answer for n. Since the original variable in the equation is k, and k=n+1, then the value of the final term you can stop on to reach your error bound will be k |
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nth term |
|
Error Bound |
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