Absolute and Conditional convergence

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How to check absolute convergence

Absolute convergence: If the absolute value \dpi{120} \sum \left | a_{n} \right |   of the series  , \dpi{120} \sum a_{n}  is convergent , then series  is called absolutely convergent.

Absolute convergence is a  stronger type of convergence.  Series that are absolutely convergent, are guaranteed to be convergent.  However, series that are convergent may or may not be absolutely convergent.

Lets look into some examples.

 

Example1: Does the following series converge absolutely.

\dpi{120} {\color{Red} \sum_{n=1}^{\infty }\frac{(-1)^{n+1}}{n^{3}}}

Solution: lets check the absolute value of this series first.

\dpi{120} \sum_{n=1}^{\infty }\left | a_{n} \right |=\sum_{n=1}^{\infty }\left |\frac{(-1)^{n+1}}{n^{3}} \right |= \sum_{n=1}^{\infty }\frac{1}{n^{3}}

Which is a P series .

Here p=3  which is greater than 1

The series is  convergent using P series test. So the given series converges absolutely

 

Example2: Check the following series for absolute convergence.

\dpi{120} {\color{Red} \sum_{n=1}^{\infty }\frac{sin(n)}{n^{2}}}

Solution:  First note that this is not alternating series but we can check its absolute convergence.

We know that sin ranges from -1 to 1

-1 ≤  Sin(n) ≤ 1

Which can be written as,  |sin(n)| ≤  1

Therefore,      \dpi{120} \left | \frac{sin(n)}{n^{2}} \right |\leq \frac{1}{n^{2}}

We know that  \dpi{120} \frac{1}{n^{2}}   is convergent by P series test as p=2 >1 .

By comparison test  given series is convergent.

Therefore the given series is absolutely convergent.

 

Conditional convergence:

When given series \dpi{120} \sum a_{n} is  convergent but  \dpi{120} \sum \left |a_{n} \right |  is divergent then  we say series is conditionally convergent.

Let\dpi{120} \sum a_{n} is a convergent series. We say that this series is Absolutely Convergent if \dpi{120} \sum \left |a_{n} \right |  is also convergent. We say the original series is Conditionally Convergent  if  \dpi{120} \sum \left |a_{n} \right |  is not convergent.

 

Example3: Check whether the following series is absolutely convergent ,  conditionally convergent or divergent.

\dpi{120} {\color{Red} \sum_{n=1}^{\infty }\frac{(-1)^{n}}{n}}
Solution:  This is an alternating harmonic series  and we use AST test to check its convergence, which is covered in previous section. Using AST we check its two conditions.

  1. \dpi{120} \lim_{n \to \infty }a_{n}=\lim_{n \to \infty }\frac{1}{n}= \frac{1}{\infty }=0
  2. This is a decreasing series because terms get decreased when denominators  get increased, which can be checked using derivative test too.

\dpi{120} (a_{n})'=\left (\frac{1}{n} \right )' = \frac{-1}{n^{2}}    which is always negative , that means this function(series) is  always decreasing.

Since both conditions  are met so this series is convergent  by AST.

Now we check its absolute convergence.

\dpi{120} \sum_{n=1}^{\infty }\left | a_{n} \right |=\sum_{n=1}^{\infty }\left |\frac{(-1)^{n}}{n} \right |= \sum_{n=1}^{\infty }\frac{1}{n}

Which is divergent being a harmonic series.

Here we saw that original series  is convergent by AST  but  it is not convergent absolutely  so this series is conditionally convergent.

 

 

Example4: Check whether the following series is absolutely convergent ,  conditionally convergent or divergent.

\dpi{120} {\color{Red} \sum_{n=1}^{\infty }(-1)^{n}\left ( n+\frac{1}{n} \right )}
Solution:  First we check  its convergence using AST . Lets check the two conditions:

1)   \dpi{120} \lim_{n \to \infty }a_{n}=\lim_{n \to \infty }\left (n+\frac{1}{n} \right )= \infty +0 =\infty

First condition is not satisfied , no need to check it further.

This series  is divergent by AST

We can also check it  is not absolutely convergent  using Limit comparison test.

So given series is divergent.

 

 

 

 

 

 

 

Practice problems:

In the following exercises determine whether the series is absolutely convergent, conditionally convergent, or divergent.

  1. \dpi{120} \sum_{n=1}^{\infty }\frac{(-1)^{n}}{2n}
  2. \dpi{120} \sum_{n=1}^{\infty }(-1)^{n}\frac{n^{2}+1}{n^{2}}
  3. \dpi{120} \sum_{n=1}^{\infty }\frac{(-1)^{n}}{2^{n}}      [ geometric series ]

 

 

 

 

 

 

Answers:

  • Conditionally convergent
  • Divergent
  • Convergent