# Consider the following convergent series.

– Determine the remainder’s upper bound in regards to n.

– Find out how many terms you need to make sure the rest is less than $1 0^{ – 3 }$.

– Identify the accurate value of the series’ lower and upper limits (ln and Un, respectively).

The main objective of this question is to find the upper and lower bound for the convergent series.

This question uses the concept of convergent series. A series is said to converge if the sequence of its cumulative sum tends to a limit. This means that when the partial sums are added to each other in the sequence of the indices, they get progressively closer to a certain number.

a) Given that:

$\space \sum_{ k = 1 }^{ \infty } \space \frac{ 1 }{ 3 ^ k }$

For the upper bound, we have:

$\space R_n \space < \space \int_{ n }^{ \infty } \frac{ 1 }{ 3 ^ x }, dx$

$\int_{ n }^{ \infty } \frac{ 1 }{ 3 ^ x } , dx \space = \space lim_{b \rightarrow \infty} \int_{ n }^{ b } \frac{ 1 }{ 3 ^ x } , dx$

$\space = \space lim_{b \rightarrow \infty} [ – \space \frac{ 1 }{ ln(3)3^b } \space + \space \frac{1}{ ln(3)3^n }]$

$\space = \space 0 \space + \space \frac{1}{ ln(3) 3^n }$

$\space = \space \frac{ 1 }{ ln(3)3^n }$

Thus, the upper bound is:

$\space = \space \frac{ 1 }{ ln(3)3^n }$

b) Given that:

$\space \sum_{ k = 1 }^{ \infty } \space \frac{ 1 }{ 3 ^ k }$

$\space R_n \space < \space 10^{ – 3 }$

Thus:

$\frac{1}{ln( 3 ) 3^n } \space < \space \frac{1}{ 10 ^3}$

$\space ln(3) \space > \space ln( 1 0 0 0) \space – \space ln ( ln ( 3 ) )$

$\space 3^n \space > \space \frac{ 1 0 0 0}{ln ( 3 )}$

$\space n \space > \space \frac{ 3 \space – \space ln(ln(3))}{ln(3)}$

Thus:

$\space n \space > \space 2 . 6 4 5$

c) We know that:

$\space S_n \space + \space \int_{ n + 1}^{ \infty } \frac{ 1 }{ 3 ^ x } , dx \space < \space S_n \space + \space \int_{ n }^{ \infty } \frac{ 1 }{ 3 ^ x } , dx$

Thus:

$\space S_n \space + \space \frac{1}{ln(3)3^{n+1}} \space + \space S \space < \space S_n \space + \space \frac{1}{ ln(3)3^n}$

## Numerical Results

The remainder’s upper bound in regards to $n$ is:

$\space = \space \frac{ 1 }{ ln(3)3^n }$

The terms needed are:

$\space n \space > \space 2 . 6 4 5$

The accurate value of the series’ lower and upper limits is:

$\space S_n \space + \space \frac{1}{ln(3)3^{n+1}} \space + \space S \space < \space S_n \space + \space \frac{1}{ ln(3)3^n}$

## Example

Determine the remainder’s upper bound in regards to $n$.

$\space \sum_{ k = 1 }^{ \infty } \space \frac{ 1 }{ 3 ^ k }$

We are given:

$\space \sum_{ k = 1 }^{ \infty } \space \frac{ 1 }{ 4 ^ k }$

For the upper bound, we have:

$\space R_n \space < \space \int_{ n }^{ \infty } \frac{ 1 }{ 4 ^ x }, dx$

$\int_{ n }^{ \infty } \frac{ 1 }{ 4 ^ x } , dx \space = \space lim_{b \rightarrow \infty} \int_{ n }^{ b } \frac{ 1 }{ 4 ^ x } , dx$

$\space = \space lim_{b \rightarrow \infty} [ – \space \frac{ 1 }{ ln(4)4^b } \space + \space \frac{1}{ ln(4)4^n }]$

$\space = \space 0 \space + \space \frac{1}{ ln(4) 4^n }$

$\space = \space \frac{ 1 }{ ln(4)4^n }$

Thus, the upper bound is:

$\space = \space \frac{ 1 }{ ln(4)4^n }$