Chapter 8.4, Problem 34E

To determine

To calculate:The given series is absolute convergent.

Answer to Problem 34E

The given series is not absolute convergent.

Explanation of Solution

Given information:

The series is

  $\frac{2}{5}+\frac{2\cdot 6}{5\cdot 8}+\frac{2\cdot 6\cdot 10}{5\cdot 8\cdot 11}+\frac{2\cdot 6\cdot 10\cdot 14}{5\cdot 8\cdot 11\cdot 14}+\mathrm{................................}$

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.

The general term of the arithmetic progression is

  $a_n=a+\left(n-1\right)d$

Where the $a$ is the first term of A.P and $d$ is the common difference of A.P

Calculation:

The series is $\frac{2}{5}+\frac{2\cdot 6}{5\cdot 8}+\frac{2\cdot 6\cdot 10}{5\cdot 8\cdot 11}+\frac{2\cdot 6\cdot 10\cdot 14}{5\cdot 8\cdot 11\cdot 14}+\mathrm{................................}$

Let the general term of the given series is $a_n$ .

Firstly observe that the series $2,6,10,14,\mathrm{........}$ and $5,8,11,14,\mathrm{..............}$ are the two different arithmetic progression in the given series.

For the arithmetic progression $2,6,10,14,\mathrm{........}$ whose general term is $b_n$ , first term is $2$ and common difference is $4$ .

Then,

  $\begin{array}{l}{b}_n=2+\left(n-1\right)4\\ =4n-2\end{array}$

For the arithmetic progression $5,8,11,14,\mathrm{..............}$ whose general term is $c_n$ , first term is $5$ and common difference is $3$ .

Then,

  $\begin{array}{l}{c}_n=5+\left(n-1\right)3\\ =3n+2\end{array}$

Now, apply ratio test to the given series

Then,

  $\begin{array}{l}\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{\frac{b_1\cdot b_2\cdot b_3\cdot \mathrm{............}{b}_n\cdot b_{n+1}}{c_1\cdot c_2\cdot c_3\cdot \mathrm{............}{c}_n\cdot c_{n+1}}}{\frac{b_1\cdot b_2\cdot b_3\cdot \mathrm{............}{b}_n}{c_1\cdot c_2\cdot c_3\cdot \mathrm{............}{c}_n}}\right|\\ =\left|\frac{b_{n+1}}{c_{n+1}}\right|\\ =\left|\frac{4\left(n+1\right)-2}{3\left(n+1\right)+2}\right|\\ =\left|\frac{4n+2}{3n+5}\right|\\ =\frac{4n+2}{3n+5}\\ \\ \underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\frac{4n+2}{3n+5}\\ =\underset{n\to \infty }{\lim }\frac{4+\frac{2}{n}}{3+\frac{5}{n}}\\ =\frac{4}{3}>1\end{array}$

Here, If $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\frac{4}{3}>1$ then the series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ is divergent.

So, the given series is divergent.

Conclusion:

The given series is not absolute convergent.