Chapter 8: Q 47. (page 670)

Find the interval of convergence for power series: $\sum_{k = 0}^{\infty} \frac{1}{k^{k}} {(x - 3)}^{k}$

Short Answer

Expert verified

The interval of convergence for power series is $R$ .

Step by step solution

Step 1. Given information.

The given power series is $\sum_{k = 0}^{\infty} \frac{1}{k^{k}} {(x - 3)}^{k}$ .

Step 2. Find the interval of convergence.

Here we use modified root test for the series. This is reasonable choice for the series because the factors of the terms of the series involves $k^{t h}$ power.

Let us take $b_{k} = \frac{1}{k^{k}} {(x - 3)}^{k}$

Thus, $\lim_{k \to \infty} \sqrt[k]{|b_{k}|} = \lim_{k \to \infty} \sqrt[k]{|\frac{1}{k^{k}} {(x - 3)}^{k}|} = \lim_{k \to \infty} \frac{1}{k} |x - 3|$

Now, we evaluating the preceding limit as $k \to \infty$ .No matter what the variable $x$ takes on the limit is zero.

That is $\lim_{k \to \infty} \frac{1}{k} |x - 3| = 0$

Therefore, by the modified root test the series converges absolutely for every value of $x$ .

Thus, the interval of the convergence is $R$ .

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Find the interval of convergence for power series: $\sum_{k = 0}^{\infty} \frac{1}{k^{k}} {(x - 3)}^{k}$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Find the interval of convergence.

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Find the interval of convergence for power series:∑k=0∞1kkx-3k

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Find the interval of convergence.

One App. One Place for Learning.

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Find the interval of convergence for power series: $\sum_{k = 0}^{\infty} \frac{1}{k^{k}} {(x - 3)}^{k}$