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Chapter 8: Q 51. (page 670)

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

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 ${鈭�}_{k = 0}^{鈭�} \frac{k!}{(k + m)! (k + m)!} x^{k}$ .

02

Step 2. Find the radius of convergence.

Let us take $b_{k} = \frac{k!}{(k + m)! (k + m)!} x^{k}$ therefore localid="1649443626355" $b_{k + 1} = \frac{(k + 1)!}{(k + 1 + m)! (k + 1 + m)!} x^{k + 1}$

Ratio for the absolute convergence is

localid="1649443468258" $\lim_{k 鈫� 鈭�} |\frac{b_{k + 1}}{b_{k}}| = \lim_{k 鈫� 鈭�} |\frac{\frac{(k + 1)!}{(k + 1 + m)! (k + 1 + m)!} x^{k + 1}}{\frac{k!}{(k + m)! (k + m)!} x^{k}}| = \lim_{k 鈫� 鈭�} |(\frac{k + 1}{(k + m) (k + m)}) x|$

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

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

So, 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

Find the interval of convergence for power series: ${鈭�}_{k = 0}^{鈭�} \frac{1}{k!} x^{k}$

Find the interval of convergence for power series: ${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} x^{2 k + 1}$

Exercise 64-68 concern with the bessel function.

What is the interval for convergence for $J_{0} (x) ?$

If a function f has a Taylor series at $x_{0}$ , what are the possibilities for the interval of convergence for that series?

Show that the series: $\ln (2) + {鈭�}_{k = 1}^{鈭�} \frac{{(- 1)}^{k - 1}}{k 2^{k}} {(x - 2)}^{k}$

from Example 3 diverges when x = 0 and converges conditionally when x = 4.

See all solutions

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