Difference between revisions of "Series Problems"

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::<span class="exam"><math>\sum_{n=1}^\infty \frac{n}{n^4+1}</math>
 
::<span class="exam"><math>\sum_{n=1}^\infty \frac{n}{n^4+1}</math>
  
== [[031_Review Part 1,_Problem_4|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 4&nbsp;</span>]] ==
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== [[Series Problems,_Problem_4|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 4&nbsp;</span>]] ==
<span class="exam"> True or false: If &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is invertible, then &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is diagonalizable.
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::<span class="exam"><math>\sum_{n=1}^\infty \frac{n^2-1}{3n^4+1}</math>
  
== [[031_Review Part 1,_Problem_5|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 5&nbsp;</span>]] ==
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== [[Series Problems,_Problem_5|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 5&nbsp;</span>]] ==
<span class="exam">True or false: If &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">B</math>&nbsp; are invertible &nbsp;<math style="vertical-align: 0px">n\times n</math>&nbsp; matrices, then so is &nbsp;<math style="vertical-align: -1px">A+B.</math>
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::<span class="exam"><math>\sum_{n=1}^\infty \frac{(-1)^{n-1}n^2}{10^n}</math>
  
== [[031_Review Part 1,_Problem_6|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 6&nbsp;</span>]] ==
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== [[Series Problems,_Problem_6|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 6&nbsp;</span>]] ==
<span class="exam"> True or false: If &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is a &nbsp;<math style="vertical-align: 0px">3\times 5</math>&nbsp; matrix and &nbsp;<math style="vertical-align: -4px">\text{dim Nul }A=2,</math>&nbsp; then &nbsp;<math style="vertical-align: 0px">A\vec{x}=\vec{b}</math>&nbsp; is consistent for all &nbsp;<math style="vertical-align: 0px">\vec{b}</math>&nbsp; in &nbsp;<math style="vertical-align: 0px">\mathbb{R}^3.</math>
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::<span class="exam"><math>\sum_{n=1}^\infty \frac{10^n}{(n+1)4^{2n+1}}
  
== [[031_Review Part 1,_Problem_7|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 7&nbsp;</span>]] ==
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== [[Series Problems,_Problem_7|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 7&nbsp;</span>]] ==
 
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::<span class="exam"><math>\sum_{n=1}^\infty \frac{(n^2+1)^{2n}}{(2n^2+1)^n}</math>
<span class="exam">True or false: Let &nbsp;<math style="vertical-align: 0px">C=AB</math>&nbsp; for &nbsp;<math style="vertical-align: 0px">4\times 4</math>&nbsp; matrices &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">B.</math>&nbsp; If &nbsp;<math style="vertical-align: 0px">C</math>&nbsp; is invertible, then &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is invertible.
 
  
 
== [[031_Review Part 1,_Problem_8|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 8&nbsp;</span>]] ==
 
== [[031_Review Part 1,_Problem_8|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 8&nbsp;</span>]] ==

Revision as of 13:34, 22 October 2017

These questions are meant to be practice problems for series.

Determine whether the series converge or diverge.

Click on the  boxed problem numbers  to go to a solution.

 Problem 1 

 Problem 2 

 Problem 3 

 Problem 4 

 Problem 5 

 Problem 6 

Failed to parse (syntax error): {\displaystyle \sum_{n=1}^\infty \frac{10^n}{(n+1)4^{2n+1}} == [[Series Problems,_Problem_7|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 7&nbsp;</span>]] == ::<span class="exam"><math>\sum_{n=1}^\infty \frac{(n^2+1)^{2n}}{(2n^2+1)^n}}

 Problem 8 

True or false: Let    be a subspace of    and    be a vector in    If    and    then  

 Problem 9 

True or false: If    is an invertible    matrix, and    and    are    matrices such that    then