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

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**.

## Expert Answer

**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:

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

\[ \Rightarrow S_{ 5 } \ \approx \ -0.28347 \ + \ 0.000002 \]

\[ \Rightarrow S_{ 5 } \ \approx \ -0.283468 \]