Chapter 8.4, Problem 42E

a.

To determine

To calculate: Whether the series is convergent.

Answer to Problem 42E

The given series is convergent.

Explanation of Solution

Given information:

  $\frac{1}{\pi }=\frac{2\sqrt{2}}{9801}{\displaystyle \sum _{n=0}^{\infty }\frac{\left(4n\right)!\left(1103+26390n\right)}{{\left(n!\right)}^4{396}^{4n}}}$

Concept Used:

Ratio Test:

(i) If $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=L<1$ , then the series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$is absolutely convergent.

(and therefore convergent)

(ii) If $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=L>1$ or $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\infty$ , then the series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$is divergent.

(iii) If $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=1$, the Ratio test is inconclusive.

Calculation:

The series is $\frac{1}{\pi }=\frac{2\sqrt{2}}{9801}{\displaystyle \sum _{n=0}^{\infty }\frac{\left(4n\right)!\left(1103+26390n\right)}{{\left(n!\right)}^4{396}^{4n}}}$

  $a_n=\frac{\left(4n\right)!\left(1103+26390n\right)}{{\left(n!\right)}^4{396}^{4n}}$

Then, $a_{n+1}$

  $a_{n+1}=\frac{\left(4\left(n+1\right)\right)!\left(1103+26390\left(n+1\right)\right)}{{\left(\left(n+1\right)!\right)}^4{396}^{4\left(n+1\right)}}$

By use of ratio test

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\frac{\left(4\left(n+1\right)\right)!\left(1103+26390\left(n+1\right)\right)}{{\left(\left(n+1\right)!\right)}^4{396}^{4\left(n+1\right)}}\times \frac{{\left(n!\right)}^4{396}^{4n}}{\left(4n\right)!\left(1103+26390n\right)}\\ =\underset{n\to \infty }{\lim }\frac{\left(4n+4\right)\left(4n+3\right)\left(4n+2\right)\left(4n+1\right)\left(1103+26390+26390n\right)}{{\left(n+1\right)}^4{\left(396\right)}^4\left(1103+26390n\right)}\\ =\underset{n\to \infty }{\lim }\frac{\left(4+\frac{4}{n}\right)\left(4+\frac{3}{n}\right)\left(4+\frac{2}{n}\right)\left(4+\frac{1}{n}\right)\left(\frac{27493}{n}+26390\right)}{{\left(1+\frac{1}{n}\right)}^4{\left(396\right)}^4\left(\frac{1103}{n}+26390\right)}\\ =\frac{4^4}{{396}^4}\\ =\frac{1}{{99}^4}\end{array}$

Means,

  $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{{99}^4}<1$

So, $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{{99}^4}<1$ , then the series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ is absolutely convergent.

Conclusion:

The given series is convergent.

b.

To determine

To calculate:The number of correct decimal places of $\pi$ by considering one term of series and two term of series.

Answer to Problem 42E

The number of correct decimal places of $\pi$ by considering one term of series and two term of series.

Explanation of Solution

Given information:

  $\frac{1}{\pi }=\frac{2\sqrt{2}}{9801}{\displaystyle \sum _{n=0}^{\infty }\frac{\left(4n\right)!\left(1103+26390n\right)}{{\left(n!\right)}^4{396}^{4n}}}$

Calculation:

The series is $\frac{1}{\pi }=\frac{2\sqrt{2}}{9801}{\displaystyle \sum _{n=0}^{\infty }\frac{\left(4n\right)!\left(1103+26390n\right)}{{\left(n!\right)}^4{396}^{4n}}}$

  $a_n=\frac{\left(4n\right)!\left(1103+26390n\right)}{{\left(n!\right)}^4{396}^{4n}}$

• For considering first term

  $n=0$

  $\begin{array}{l}{a}_0=\frac{\left(4\times 0\right)!\left(1103+26390\left(0\right)\right)}{{\left(0!\right)}^4{396}^{4\times 0}}\\ =1103\end{array}$

Then

  $\begin{array}{l}\frac{1}{\pi }=\frac{2\sqrt{2}}{9801}{a}_0\\ =\frac{2\sqrt{2}}{9801}\times 1103\\ =0.318309878\end{array}$

Then,

  $\begin{array}{l}\frac{1}{\pi }=0.318309886\\ \pi =\frac{1}{0.318309886}\\ \text{}=3.14159273\end{array}$

And we know the value of

  $\pi =3.14159265$

From this it can be seen that the given series gives 6 decimal place correct for considering first term.

  $\pi =3.141592$

• For considering first two terms

  $n=0\text{and}n=1$

  $\begin{array}{l}\frac{1}{\pi }=\frac{2\sqrt{2}}{9801}{\displaystyle \sum _{n=0}^1{a}_n}\\ =\frac{2\sqrt{2}}{9801}\left(a_0+a_1\right)\end{array}$

For $a_1$

  $\begin{array}{l}{a}_1=\frac{\left(4\times 1\right)!\left(1103+26390\left(1\right)\right)}{{\left(1!\right)}^4{396}^{4\times 1}}\\ =\frac{24\times 27493}{{396}^4}\\ =2.68319743\times {10}^{-5}\end{array}$

Put value of $a_{1}and{a}_0$ in the equation

  $\begin{array}{l}\frac{1}{\pi }=\frac{2\sqrt{2}}{9801}\left(a_0+a_1\right)\\ =\frac{2\sqrt{2}}{9801}(1103+2.683197435\times {10}^{-5})\\ =0.318309886\\ \text{Then}\\ \pi =\frac{1}{0.318309886}\\ =3.141592654\end{array}$

Conclusion:

From this it can be seen that the given series gives 6 decimal place correct for considering first term.

  $\pi =3.141592654$