Calculus 2 : Convergence and Divergence

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #122 : Series In Calculus

Find the radius of convergence of 

\displaystyle \sum_{n=0}^{\infty} \frac{x^{5n}}{3^n}.

Possible Answers:

\displaystyle \infty

\displaystyle \frac{1}{3^{\frac{1}{5}}}

\displaystyle 3^{\frac{1}{5}}

\displaystyle 5^{\frac{1}{3}}

Correct answer:

\displaystyle 3^{\frac{1}{5}}

Explanation:

We can find the radius of convergence of

\displaystyle \sum_{n=0}^{\infty} \frac{x^{5n}}{3^n}

using the ratio test:

\displaystyle \lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n} \right |=\lim_{n\to \infty} \left|\frac{x^{5n+5}}{3^{n+1}}\frac{3^n}{x^{5n}} \right |=\lim_{n\to \infty} \left|\frac{x^{5}}{3} \right |=\left|\frac{x^5}{3}\right|< 1

which means that \displaystyle |x|< 3^{\frac{1}{5}}

So that \displaystyle R=3^{\frac{1}{5}} is the radius of convergence.

 

Example Question #123 : Series In Calculus

Determine whether the series is convergent or divergent:

\displaystyle \sum_{n=0}^{\infty }\frac{n!}{3^{n+2}}

Possible Answers:

The series is divergent

The series is (absolutely) convergent

The series is conditionally convergent

The series may be (absolutely) convergent, divergent, or conditionally convergent

Correct answer:

The series is divergent

Explanation:

To determine whether the series given converges or diverges, we must use the Ratio Test, which states that for 

\displaystyle L=\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_{n}}\right |, if \displaystyle L< 1, then series is (absolutely) convergent; if \displaystyle L=1, then the series may be (absolutely) convergent, divergent, or conditionally convergent; and if \displaystyle L>1, then the series is divergent. 

So, we must evaluate the limit as given by the above formula:

\displaystyle L=\lim_{n\rightarrow \infty } \frac{(n+1)!}{3^{n+3}}\cdot \frac{3^{n+2}}{(n)!}=\lim_{n\rightarrow \infty } \frac{n+1}{3}=\infty >1

Thus, the series is divergent. 

(The limit was solved by \displaystyle \lim_{n\rightarrow \infty } \frac{n}{n}(\frac{1+n^{-1}}{3n^{-1}})=\infty. As \displaystyle n approaches infinity, the negative terms go to zero, which makes zero in the denominator and \displaystyle 1 in the numerator, equalling infinity.)

Example Question #124 : Series In Calculus

Determine if the series converges 

\displaystyle \sum_{n=0}^{\infty }\frac{3^n}{2^{2n+1}}

 

Possible Answers:

Cannot be determined

Series diverges

Series converges

Correct answer:

Series converges

Explanation:

In order to test the convergence of the series, we use the ratio test.

\displaystyle L=\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |

If...

\displaystyle L< 1 the series is absolutely convergent

\displaystyle L=1 the convergence is undetermined

\displaystyle L>1 the series diverges

For this problem

\displaystyle L=\lim_{n\rightarrow \infty }\left | \frac{\frac{3^{n+1}}{2^{2(n+1)+1}}}{\frac{3^n}{2^{2n+1}}} \right |

\displaystyle =\lim_{n\rightarrow \infty }\left | \frac{3^{n+1}}{2^{2(n+1)+1} } \cdot \frac{2^{2n+1}}{3^n} \right |

\displaystyle =\lim_{n\rightarrow \infty }\left | \frac{3\cdot3^{n}}{2^2\cdot2^{2n+1} } \cdot \frac{2^{2n+1}}{3^n} \right |

\displaystyle =\lim_{n\rightarrow \infty }\left | \frac{3}{2^2} } \right |

\displaystyle =\frac{3}{4}

And since 

\displaystyle L< 1 the series absolutely converges.

Example Question #131 : Series In Calculus

Use the ratio test to figure out if the series 

\displaystyle \sum_{n=1}^\infty \frac{1}{n^2+n}

converges.

Possible Answers:

It converges

it diverges to infinity

Correct answer:

It converges

Explanation:

an easy sequence to compare \displaystyle \frac{1}{n^2+n} to is \displaystyle \frac{1}{n^2}, which we know converges.

\displaystyle \frac{\frac{1}{n^2}}{\frac{1}{n^2+n}} = \frac{n^2+n}{n^2}

We need to find \displaystyle \lim_{n \to \infty} \frac{n^2+n}{n^2} = \lim_{n \to \infty} 1+\frac{1}{n}

\displaystyle \frac{1}{n} goes to zero so this limit is \displaystyle 1. Since the limit of the ratio is 1 and \displaystyle \sum \frac{1}{n^2} converges, \displaystyle \sum \frac{1}{n^2+n} must converge also.

Example Question #92 : Convergence And Divergence

Can we use the Ratio Test to test the convergence/divergence of the infinite series \displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^3} ?

Possible Answers:

Yes, and the series converges

No, the ratio test fails here.

Yes, and the series diverges

Correct answer:

No, the ratio test fails here.

Explanation:

To use the ratio test, we must first evaluate

\displaystyle \lim_{k \to \infty} |\frac{a_{k+1}}{a_k}|

\displaystyle =\lim_{k \to \infty} |\frac{\frac{(-1)^{k+1}}{(k+1)^3}}{\frac{(-1)^k}{k^3}}|

\displaystyle =\lim_{k \to \infty} |\frac{(-1)^{k+1}}{(k+1)^{3}} \times \frac{k^3}{(-1)^k}|

\displaystyle = \lim_{k \to \infty} | -1 \times \frac{k^3}{(k+1)^3}|

\displaystyle = \lim_{k \to \infty} \frac{k^3}{(k+1)^3}

\displaystyle =1

Since the result of the limit is \displaystyle 1, the Ratio Test fails, and therefore we cannot use it in testing this series for convergence/divergence.

Example Question #132 : Series In Calculus

Determine whether the series converges or diverges:

\displaystyle \sum_{n=0}^{\infty} \frac{(n+2)!}{5^n}

Possible Answers:

The series diverges

The series converges

The series may be convergent, divergent, or conditionally convergent

The series is conditionally convergent

Correct answer:

The series diverges

Explanation:

To determine whether the series converges or diverges, we must use the ratio test, which states that for the given series \displaystyle \sum_{n=0}^{\infty} a_{n}, and 

  \displaystyle L=\lim_{n\rightarrow \infty} \left |\frac{ a_{n+1}}{a_{n}}\right |, when L is equal to 1, then the series may be convergent, divergent, or conditionally convergent; when L is less than 1 then the series is convergent; and when L is greater than 1 the series is divergent.

Using the above test, we get

\displaystyle L=\lim_{n\rightarrow \infty} \left | \frac{(n+3)!}{5^{n+1}} \cdot \frac{5^n}{(n+2)!} \right |=\lim_{n\rightarrow \infty}\left | \frac{n+3}{5^1}\right |=\infty>1

Because L is greater than 1, the series diverges.

Example Question #97 : Ratio Test

\displaystyle \sum_{n=0}^{\infty } \frac{n^{2}}{3^{n}}

Determine the convergence of the series using the Ratio Test.

Possible Answers:

Cannot be determined

Series converges

Series diverges

Correct answer:

Series converges

Explanation:

The ratio test is defined as follows

\displaystyle \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |

where \displaystyle a_n is the n-th term of the series.

if the limit

  • is greater than \displaystyle 1, the series diverges
  • is less than \displaystyle 1, the series converges
  • equal to \displaystyle 1, the test is inconclusive

Finding the limit for the terms in our series,

\displaystyle \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |=\lim_{n \to \infty} \left | \frac{\frac{(n+1)^2}{3^{n+1}}}{\frac{n^2}{3^n}} \right |

\displaystyle =\lim_{n \to \infty}\left | \frac{(n^2+2n+1)3^n}{(n^2)3^{n+1}} \right |

\displaystyle =\frac{1}{3}\lim_{n \to \infty}\left | \frac{n^2+2n+1}{n^2} \right |

\displaystyle =\frac{1}{3}*1

\displaystyle =\frac{1}{3}

Because the limit is less than \displaystyle 1,

the series converges.

Example Question #98 : Ratio Test

\displaystyle \sum_{n=0}^{\infty}\frac{(ln\,n)3^n}{2^{2n+1}}

Determine the convergence of the series using the Ratio Test.

Possible Answers:

Cannot be determined

Series diverges

Series converges

Correct answer:

Series converges

Explanation:

The ratio test is defined as follows

\displaystyle \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |

where \displaystyle a_n is the n-th term of the series.

if the limit

  • is greater than \displaystyle 1, the series diverges
  • is less than \displaystyle 1, the series converges
  • equal to \displaystyle 1, the test is inconclusive

Finding the limit for the terms in our series,

\displaystyle \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |=\lim_{n \to \infty} \left | \frac{\frac{(ln\,n+1)3^{n+1}}{2^{2(n+1)+1}}}{\frac{(ln\,n)3^n}{2^(2n+1)}} \right |

\displaystyle =\lim_{n \to \infty}\left |\frac{(ln\,n+1)2^{2n+1}*3^{n+1}}{(ln\,n)2^{2n+3}*3^n} \right |

\displaystyle =\frac{3}{2^2}\lim_{n \to \infty}\left |\frac{ln\,n+1}{ln\,n} \right |

\displaystyle =\frac{3}{4}*1

\displaystyle =\frac{3}{4}

Because the limit is less than \displaystyle 1,

the series converges.

Example Question #99 : Ratio Test

\displaystyle \sum_{n=1}^{\infty}\frac{n!}{4^n}

Possible Answers:

Series converges

Series diverges

Cannot be determined

Correct answer:

Series diverges

Explanation:

The ratio test is defined as follows

\displaystyle \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |

where \displaystyle a_n is the n-th term of the series.

if the limit

  • is greater than \displaystyle 1, the series diverges
  • is less than \displaystyle 1, the series converges
  • equal to \displaystyle 1, the test is inconclusive

Finding the limit for the terms in our series,

\displaystyle \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |=\lim_{n \to \infty} \left | \frac{\frac{(n+1)!}{4^{n+1}}}{\frac{n!}{4^n}} \right |

\displaystyle =\lim_{n \to \infty}\left |\frac{(n+1)n!4^n}{n!4^{n+1}} \right |

\displaystyle =\frac{1}{4}\lim_{n \to \infty}\left |n+1\right |

\displaystyle =\frac{1}{4}*\infty

\displaystyle =\infty

Because the limit is greater than \displaystyle 1,

the series diverges.

Example Question #100 : Ratio Test

\displaystyle \sum_{n=1}^{\infty}\frac{n^3}{n!2^n}

Determine the convergence of the series.

Possible Answers:

Series diverges

Series converges

Cannot be determined

Correct answer:

Series converges

Explanation:

The ratio test is defined as follows

\displaystyle \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |

where \displaystyle a_n is the n-th term of the series.

if the limit

  • is greater than \displaystyle 1, the series diverges
  • is less than \displaystyle 1, the series converges
  • equal to \displaystyle 1, the test is inconclusive

Finding the limit for the terms in our series,

\displaystyle \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |=\lim_{n \to \infty} \left | \frac{\frac{(n+1)^3}{(n+1)!2^{n+1}}}{\frac{n^3}{n!2^n}} \right |

\displaystyle =\lim_{n \to \infty}\left | \frac{n!(n+1)^3*2^n}{(n+1)n!n^3*2^{n+1}} \right |

\displaystyle =\frac{1}{2}\lim_{n \to \infty}\left | \frac{(n+1)^2}{n^3}\right |

\displaystyle =\frac{1}{2}*0

\displaystyle =0

Because the limit is less than \displaystyle 1,

the series converges.

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