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Chapter 8: Q 52. (page 670)

Find the radius of convergence for the given series: ${鈭�}_{k = 0}^{鈭�} \frac{{(k!)}^{m}}{(m k)!} x^{k}$

Short Answer

Expert verified

The radius of convergence for the series is $R$ -

Step by step solution

01

Step 1. Given information.

The given power series is ${鈭�}_{k = 0}^{鈭�} \frac{{(k!)}^{m}}{(m k)!} x^{k}$ .

02

Step 2. Find the radius of convergence.

Let us take $b_{k} = \frac{{(k!)}^{m}}{(m k)!} x^{k}$ therefore $b_{k + 1} = \frac{{((k + 1)!)}^{m}}{(m (k + 1))!} x^{k + 1}$

Ratio for the absolute convergence is

$\lim_{k 鈫� 鈭�} |\frac{b_{k + 1}}{b_{k}}| = \lim_{k 鈫� 鈭�} |\frac{\frac{{((k + 1)!)}^{m}}{(m (k + 1))!} x^{k + 1}}{\frac{{(k!)}^{m}}{(m k)!} x^{k}}| = \lim_{k 鈫� 鈭�} |(\frac{{(k + 1)}^{m} x}{(m k + m) (m k + m - 1) . . . . . . (m k + 1)}) x|$

Since for $k 鈫� 鈭�$ the limit of the zero irrespective of the value of variable.

Thus,

$\lim_{k 鈫� 鈭�} |(\frac{{(k + 1)}^{m} x}{(m k + m) (m k + m - 1) . . . . . . (m k + 1)}) x| = 0$

Hence, by the ratio test the series converges absolutely for every value of $x$ .

Therefore, the radius of the convergence for the series is $R$ .

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Most popular questions from this chapter

What is $x_{0}$ if the interval of convergence for the power series ${鈭�}_{k = 0}^{鈭�} 鈥� a_{k} {(x 鈭� x_{0})}^{k} is (2, 10] ?$

Find the interval of convergence for power series: ${鈭�}_{k = 0}^{鈭�} \frac{\sqrt{k}}{k!} {(4 x + 7)}^{k}$

Use an appropriate Maclaurin series to find the values of the series in Exercises 17鈥�22.

${鈭�}_{k = 0}^{鈭�} {(- \frac{1}{蟺})}^{k}$

If $f (x) = 4 x^{3} - 5 x^{2} + 6 x + 1 a n d P_{3} (x)$ is the third Taylor polynomial for f at 鈭�1, what is the third remainder $R_{3} (x)$ ? What is $R_{4} (x)$ ? (Hint: You can answer this question without finding any derivatives.)

In exercises 59-62 concern the binomial series to find the maclaurin series for the given function .

$\frac{1}{\sqrt{1 + x}}$

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