 # Approximate the sum of the series correct to four decimal places.

$\boldsymbol{ \sum_{ n }^{ \infty } n \ = \ \dfrac{ 1 ( -1 )^{ n } }{ 3^{ n } n ! } }$

This question aims to develop a basic understanding of summation expressions.

A summation expression is a type of expression used to describe a series in a compact form. To find the values of such expressions we may need to solve the series for the unknowns. The solution to such a question can be very complex and time taking. If the expression is simple, one may use the manual method to solve it.

In the real world, such expressions are extensively used in computer science. The approximations of such expressions can yield significant gains in the performance of computation algorithms both in terms of space and time.

Given:

$\sum_{ n }^{ \infty } n \ = \ \dfrac{ 1 ( -1 )^{ n } }{ 3^{ n } n ! }$

We can immediately see that it is an alternating type of series. This means that the value of the term in this series alternates successfully between positive and negative values.

In the case of the alternating type of series, we can neglect the first term. This assumption yields the following expression:

$| R_{ n } | \ \le \ b_{ n + 1 } \ = \ \dfrac{ 1 }{ 3^{ n + 1 } ( n + 1 ) ! \ < \ 0.00001 }$

Now the above inequality can be very complex and tough to solve using empirical methods. So, we can use a simpler graphical or manual method to evaluate different values of the above term.

At $n \ = 4 \$:

$\dfrac{ 1 }{ 3^{ 4 + 1 } ( 4 + 1 ) ! } \ = \ \dfrac{ 1 }{ 3^{ 5 } ( 5 ) ! \ \approx \ 0.00003 } \ > \ 0.00001$

At $n \ = 5 \$:

$\dfrac{ 1 }{ 3^{ 5 + 1 } ( 5 + 1 ) ! } \ = \ \dfrac{ 1 }{ 3^{ 6 } ( 6 ) ! \ \approx \ 0.000002 } \ < \ 0.00001$

Which is the required accuracy. Therefore we can conclude that a minimum of 5 terms will be required to achieve the desired error constraint.

The sum of the first 5 terms can be calculated as:

$S_{ 5 } \ = \ \sum_{ n = 1 }^{ 5 } n \ = \ \dfrac{ 1 ( -1 )^{ n } }{ 3^{ n } n ! }$

$\Rightarrow S_{ 5 } \ = \ \dfrac{ 1 ( -1 )^{ 1 } }{ 3^{ 1 } 1 ! } + \dfrac{ 1 ( -1 )^{ 2 } }{ 3^{ 2 } 2 ! } + \dfrac{ 1 ( -1 )^{ 3 } }{ 3^{ 3 } 3 ! } + \dfrac{ 1 ( -1 )^{ 4 } }{ 3^{ 4 } 4 ! } + \dfrac{ 1 ( -1 )^{ 5 } }{ 3^{ 5 } 5 ! }$

$\Rightarrow S_{ 5 } \ \approx \ -0.28347$

## Numerical Result

$S_{ 5 } \ \approx \ -0.28347$

## Example

Calculate the result accurately up to the 5th decimal place (0.000001).

At $n \ = 5 \$:

$\dfrac{ 1 }{ 3^{ 5 + 1 } ( 5 + 1 ) ! } \ = \ \dfrac{ 1 }{ 3^{ 6 } ( 6 ) ! \ \approx \ 0.000002 } \ > \ 0.000001$

At $n \ = 6 \$:

$\dfrac{ 1 }{ 3^{ 6 + 1 } ( 6 + 1 ) ! } \ = \ \dfrac{ 1 }{ 3^{ 7 } ( 7 ) ! \ \approx \ 0.00000009 } \ < \ 0.000001$

Which is the required accuracy. Therefore we can conclude that a minimum of 6 terms will be required to achieve the desired error constraint.

The sum of the first 6 terms can be calculated as:

$S_{ 6 } \ = \ \sum_{ n = 1 }^{ 6 } n \ = \ \dfrac{ 1 ( -1 )^{ n } }{ 3^{ n } n ! }$

$\Rightarrow S_{ 5 } \ \approx \ -0.28347 \ + \ 0.000002$

$\Rightarrow S_{ 5 } \ \approx \ -0.283468$

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