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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


Test for Diverg­enc­e(TFD)
You absolutely cannot determine if a series is convergent from this test.
If limit of series ≠0 or ∞
Integral Test
If integral of series <∞
If integral of series =∞
Ratio Test
Conver­ges­/Co­nverges Absolutely
If limit of 0≤|(ak+1) /(ak)|<1
If limit >1
If limit =1
Root Test
Conver­ges­/Co­nverges Absolutely
If 0≤limit of the kth root of |ak|<1
If limit of the kth root of |ak|>1
If limit of kth root of |ak|=1
Direct Comparison Test(CT)
If ∑bk converges AND bk is the larger of the two functions
If ∑bk diverges AND bk is smaller of two functions
Limit Comparison Test(LCT)
If bk converges AND limit of 0<(ak)/(bk)<∞
If bk AND limit of 0<(ak)/(bk)<∞
Altern­ating Series Test(AST)
If all 3 conditions for AST are met
If limit condition fails, ∑ak is immedi­ately divergent by TFD

Properties of Special Series

Geometric Series
If Absolute Value of r<1. Converges at S=(first term)/­(1-r)
If Absolute Value of r≥1
If p>1
If p≤1
Value that the limit of the remaining terms approach
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

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. |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 kth power, you can use the root test

Integral Test

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

1. bk>0
2. bk≥bk+1
3. limit of bk=0

* If ∑bk fits all three condit­ions, ∑ak convergent by AST
* If 3rd condition fails, ∑ak is divergent by TFD
* If series contains (-m)k, pull (-1)k out and keep mk in bk

Integral Remainder

Known nth 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
nth term
Error Bound


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