General RulesTelescoping 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 
TestsTest 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≤(ak+1 ) /(ak )<1   Diverges  If limit >1   Inconclusive  If limit =1  Root Test  Converges/Converges Absolutely  If 0≤limit of the k^{th} root of ak <1   Diverges  If limit of the k^{th} root of ak >1   Inconclusive  If limit of k^{th} root of ak =1  Direct Comparison Test(CT)  Converges  If ∑bk converges AND bk is the larger of the two functions   Diverges  If ∑bk diverges AND bk is smaller of two functions  Limit Comparison Test(LCT)  Converges  If bk converges AND limit of 0<(ak )/(bk )<∞   Diverges  If bk AND limit of 0<(ak )/(bk )<∞  Alternating Series Test(AST)  Converges  If all 3 conditions for AST are met   Diverges  If limit condition fails, ∑ak is immediately divergent by TFD 
  Properties of Special SeriesGeometric Series  Converges  If Absolute Value of r<1. Converges at S=(first term)/(1r)   Diverges  If Absolute Value of r≥1  PSeries  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 ∑ak  is convergent   Conditional Convergence  If and only if ∑ak is convergent, but ∑ak  is divergent 
When to Use TestsProperties  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 pseries. ak  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 k^{th} power, you can use the root test 
  Integral TestConditions:
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 ak 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 ∑ak converges 
Alternating Series TestConditions:
1. bk >0
2. bk ≥bk+1
3. limit of bk =0
* If ∑bk fits all three conditions, ∑ak convergent by AST
* If 3^{rd} condition fails, ∑ak is divergent by TFD
* If series contains (m)^{k}, pull (1)^{k} out and keep m^{k} in bk 
Integral RemainderKnown n^{th} Value  Solving for Sn (sum of the series approximation)  Plug n into the series   Solving for Rn (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 RemainderKnown Error Bound  Set error bound less than bn+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, bn+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   n^{th} term  bn+1  Error Bound 

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anthonyackeret, 20:07 5 Apr 20
looks like a good summary page
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