Alternating Series Test

 

Any series with alternate signs is called alternating series  and can be represented as

\dpi{150} \sum_{n=1}^{\infty }(-1)^n a_n

 

 

To test the convergence of this type of series, we use the alternating series test(AST). This test says:

 

For a given series,  \dpi{100} \sum_{n=1}^{\infty}(-1)^n {a_n}  , where  \dpi{150} {a_n}\geq 0  for all n and if

 

i )  \dpi{150} {\color{DarkBlue} \sum_{m}^{n}}
ii)  If   \dpi{150} {a_n} is decreasing sequence  i.e   for all  n, \dpi{150} \large {\color{DarkBlue} a_{n+1} \leq a_{n} }
Then given series is convergent.

 

Example: Check the convergence of the following alternating series.

\dpi{150} {\color{Black} {}\sum_{ n=1}^{\infty}(-1)^{n-1} \frac{n^{2}}{n^{3}+1}}

 

Solution: Here we check the two conditions for the alternating series test(AST).

i) \dpi{120} \bg_white \fn_jvn \large {lim_n}_\rightarrow_{\infty}{a_n} = {lim_n}_\rightarrow_{\infty}{}\frac{n^2}{n^3+1}

\dpi{120} = {lim_n}_\rightarrow_{\infty}{}\frac{\verb|\cancel{n^2}}{\verb|\cancel{n^2}({n+1})}

\dpi{120} = {lim_n}_\rightarrow_{\infty}{}\frac{1}{({n+1})} = {}\frac{1}{\infty} = 0

 

ii)  Clearly, this is a decreasing sequence because the denominator is much larger than numerator and denominator increase largely with the increase  in the value of n. Therefore terms of a sequence  are decrease as in increase.

Since both conditions of AST are met so this series is convergent!

 

Example: Determine if the following series is convergent or divergent.

\dpi{120} {\color{Red} {}\sum_{n=1}^{\infty }(-1)^n {}\frac{n^2}{n^3+1}}

This is alternating series and we use AST  to check its convergence.

We check the two conditions one by one.

i)     \dpi{120} {lim_n}_\rightarrow_{\infty}{a_n} = {lim_n}_\rightarrow_{\infty}{}\frac{n^2}{n^2+3}

ii)    \dpi{120} = {lim_n}_\rightarrow_{\infty}{}\frac{\cancel{n^2}}{\cancel{n^2}(1+{}\frac{3}{n^{2}})}

In the above step, we can divide numerator and denominator by  \dpi{150} n^{2}   instead of factoring out.

\dpi{120} = {lim_n}_\rightarrow_{\infty}{}\frac{1}{(1+{}\frac{3}{n^{2}})}

\dpi{120} = \frac{1}{(1+{}\frac{3}{n^{2}})} = \frac{1}{(1+0)} = 1

Here we got the limit as 1. The first condition is not met so any need to check the second condition.

This series is not convergent.

 

Example: Determine if the following series is convergent or divergent.

\dpi{150} {\color{Red} {}\sum_{n=2}^{\infty }{\frac{\cos(n\pi)}{\sqrt{n}}}}

 

Solution: You will be wondering how come this is an alternating series. Let’s see,

We know that

\dpi{120} \cos(\pi) = -1, \cos(2\pi) = 1, \cos(3\pi) = -1……….so on

 

so we can write   \dpi{120} \cos(n\pi) =(-1)^{n}        

So the given series is actually an alternating series.

\dpi{120} \sum_{n=2}^{\infty }{\frac{\cos(n\pi)}{\sqrt{n}}}  \dpi{120} = \sum_{n=2}^{\infty }{\frac{(-1)^n}{\sqrt{n}}}

 

Checking  the two conditions for AST

i)  \dpi{120} {lim_n}_\rightarrow_{\infty}{a_n} = {lim_n}_\rightarrow_{\infty}{}\frac{1}{\sqrt{n}}

\dpi{120} =\frac{1}{\infty} = 0

ii)  \dpi{100} a_{n} = \frac{1}{\sqrt{n}}

  which is clearly a decreasing sequence as its denominator get larger with the increase in n.

Though it can also be checked using a derivative test.

We know that function is decreasing when the first derivative is negative.

Here  \dpi{120} (a_{n})^{'} = \frac{-1}{2 \sqrt{n}} < 0

So this series is decreasing one.

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

 

 

Practice problems:Check the convergence of the following series :

1)   \sum_{n=1}^{\infty }{\frac{(-1)^{n-1}}{2n+1}}

2)   \sum_{n=1}^{\infty }(-1)^{n}{\frac{3{n}^{3}}{n^{3}+3 }}

3)  \sum_{n=1}^{\infty }{\frac{ \cos(n\pi)}{n^{2} }}

Answers

1) Convergent

2) Divergent

3) Convergent

 

 

 

 

Leave a comment