Infinite series, Divergence tests + formulas

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Last updated 12:57 AM on 7/3/26
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33 Terms

1
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when does an infinite series diverge

-when limit does not exist
-if limit is infinite

2
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when do you use telescoping series

when consecutive terms cancel, usually requires partial fractions

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first step for telescoping series

-partial fractions if necessary

-find the first couple terms and Sn

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how to finish telescoping series

find Sn and compute the limit

5
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general geometric series formula

<p></p>
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when does geometric series converge

|r|<1

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formula for sum of geometric series

a/1-r

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what if for geometric series |r|≥1

series diverges

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Divergence test

if Σan is convergent then lim an =0;

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why is it called divergence test

because it tells you if limit an does NOT equal 0 then Σan is DIVERGENT

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using divergence test, what is
lim an=0

Nothing, it is inconclusive

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can divergence test prove convergence?

no, only divergence

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conditions for integral test
let an=f(n)

f is
-positive
-decreasing
-continuous
for all x≥1

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<p>if this converges then  Σa<sub>n</sub> ___?</p>

if this converges then Σan ___?

converges

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<p>if this diverges then  Σa<sub>n</sub> ___?</p>

if this diverges then Σan ___?

diverges

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general p-series formula

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if p>1 for p-series

it converges, otherwise it diverges

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conv/div?
Σ1/n

diverge, because using the p-series, it must be p >1 and here p=1

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conv/div
Σ1/√n

diverge, because using p-series it must be p>1 and here p=1/2

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conv/div?
Σ1/n2

converge, because using p-series p>1 and here p=2

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conditions for direct comparison test

Σan,Σbn are positive series

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Direct comparison test
if Σbn converges then Σan

converges provided 0≤an≤bn

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Direct comparison test
if Σan diverges then Σbn

diverges provided 0≤an≤bn

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what series are mostly used in comparison tests?

p-series and geometric

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limit comparison test conditions

Σan,Σbn are positive series let it be (photo)

<p>Σa<sub>n</sub>,Σb<sub>n</sub> are positive series let it be (photo)</p>
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case 1 for limit comparison

if 0<L<∞, then Σan,Σbn both converge or both diverge

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case 2 for limit comparison

if L=0 and Σbn converges, then Σan also converges

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case 3 for limit comparison

if L=∞ and Σbn diverges, then Σan diverges

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Theorem 1 (Limits of sequences

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Theorem 2 (squeeze theorem for sequences)

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theorem 3 (continuous functions)

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Theorem 4 (Bounded Monotonic Sequences)

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33
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