Chapter 8: Q 52. (page 670)

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

Short Answer

Expert verified

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

Step by step solution

Step 1. Given information.

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

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

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

Since for $k \to \infty$ the limit of the zero irrespective of the value of variable.

Thus,

$\lim_{k \to \infty} |(\frac{{(k + 1)}^{m} x}{(m k + m) (m k + m - 1) . . . . . . (m k + 1)}) x| = 0$

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

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Find the radius of convergence.

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Step by step solution

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One App. One Place for Learning.

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