Chapter 8.4, Problem 40E

To determine

To calculate:The positive interger value of $k$ for which the series is convergent.

Answer to Problem 40E

The positive interger value is $k\ge 2$ for which the series is convergent.

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{n=\infty }\frac{{\left(n!\right)}^2}{\left(kn\right)!}}$ is convergent.

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 ${\displaystyle \sum _{n=1}^{n=\infty }\frac{{\left(n!\right)}^2}{\left(kn\right)!}}$

  $a_n=\frac{{\left(n!\right)}^2}{\left(kn\right)!}$

Then,

  $\begin{array}{l}{a}_{n+1}=\frac{{\left((n+1)!\right)}^2}{\left(k(n+1)\right)!}\\ =\frac{{\left(\left(n!\right)\left(n+1\right)\right)}^2}{\left(kn+k\right)!}\\ =\frac{{\left(n!\right)}^2{\left(n+1\right)}^2}{\left(kn+k\right)!}\\ a_{n+1}=\frac{{\left(n!\right)}^2{\left(n+1\right)}^2}{\left(kn+k\right)!}\end{array}$

By use of ratio test

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\left(\frac{{\left(n!\right)}^2{\left(n+1\right)}^2}{\left(kn+k\right)!}\times \frac{\left(kn\right)!}{{\left(n!\right)}^2}\right)\\ =\underset{n\to \infty }{\lim }\frac{\left(kn\right)!{\left(n+1\right)}^2}{\left(kn+k\right)!}\\ =\underset{n\to \infty }{\lim }\frac{\left(kn\right)!{\left(n+1\right)}^2}{\left(kn\right)!\left(kn+1\right)\left(kn+2\right)\mathrm{...........}\left(kn+k\right)}\\ =\underset{n\to \infty }{\lim }\frac{{\left(n+1\right)}^2}{\left(kn+1\right)\left(kn+2\right)\mathrm{...........}\left(kn+k\right)}\end{array}$

Now,

If $k=1$ , then

  $\underset{n\to \infty }{\lim }\frac{{\left(n+1\right)}^2}{\left(n+1\right)}=\infty$

Therefore, for $k=1$ , the series is divergent.

If $k=2$ , then

  $\underset{n\to \infty }{\lim }\frac{{\left(n+1\right)}^2}{\left(2n+1\right)\left(2n+2\right)}=\frac{1}{4}$

Therefore, for $k=2$ , the series is convergent.

For $k>2$ , the value of $\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}$ always rational number which is less than $1$ .

And the $\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=0$ , when $k\to \infty$

Therefore the series converges for all $k\ge 2$

Conclusion:

The positive interger value is $k\ge 2$ for which the series is convergent.