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≤(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 Series
Geometric 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 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 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 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 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 Test
Conditions:
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 Remainder
Known 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 Remainder
Known 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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