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Calculus 2 Review Sheet

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 kth root of |a
k
|<1
 
Diverges
If limit of the kth root of |a
k
|>1
 
Inconc­lusive
If limit of kth 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 ∑a
k
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 3rd condition fails, ∑a
k
is divergent by TFD
* If series contains (-m)k, pull (-1)k out and keep mk 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
 
nth term
b
n+1
Error Bound
                           
 

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looks like a good summary page

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