Math / Calculus

QuestionA) 0 of convergence of the series $\sum_{k=0}^{\infty} \frac{(x-1)^{k}}{k!}$ is equal to B) 1 10) The Maclaurin C) $\frac{1}{2}$ D) $\infty$

Studdy Solution

STEP 1

1. The series in question is $\sum_{k=0}^{\infty} \frac{(x-1)^{k}}{k!}$ , which appears to be a power series.
2. We need to determine the radius of convergence for this power series.
3. The radius of convergence $R$ of a power series $\sum_{k=0}^{\infty} a_k (x-c)^k$ can be found using the formula $\frac{1}{R} = \limsup_{k \to \infty} \sqrt[k]{|a_k|}$ .

STEP 2

1. Identify the general term $a_k$ of the series.
2. Use the formula for the radius of convergence to find $R$ .
3. Evaluate the limit to determine the radius of convergence.

STEP 3

Identify the general term $a_k$ of the series.
$a_k = \frac{1}{k!}$

STEP 4

Apply the formula for the radius of convergence $\frac{1}{R} = \limsup_{k \to \infty} \sqrt[k]{|a_k|}$ .
$\frac{1}{R} = \limsup_{k \to \infty} \sqrt[k]{\left|\frac{1}{k!}\right|}$

STEP 5

Simplify the limit expression $\sqrt[k]{\frac{1}{k!}}$ .
$\sqrt[k]{\frac{1}{k!}} = \frac{1}{\sqrt[k]{k!}}$

STEP 6

Evaluate the limit $\lim_{k \to \infty} \sqrt[k]{k!}$ . Using Stirling's approximation $k! \approx \sqrt{2\pi k} \left(\frac{k}{e}\right)^k$ , we have:
$\sqrt[k]{k!} \approx \sqrt[k]{\sqrt{2\pi k} \left(\frac{k}{e}\right)^k}$

STEP 7

Simplify $\sqrt[k]{\sqrt{2\pi k} \left(\frac{k}{e}\right)^k}$ .
$\sqrt[k]{\sqrt{2\pi k} \left(\frac{k}{e}\right)^k} = \sqrt[k]{\sqrt{2\pi k}} \cdot \sqrt[k]{\left(\frac{k}{e}\right)^k}$
$\sqrt[k]{\sqrt{2\pi k}} \to 1 \quad \text{as} \quad k \to \infty$
$\sqrt[k]{\left(\frac{k}{e}\right)^k} = \frac{k}{e}$

STEP 8

Combine the results to find the limit.
$\lim_{k \to \infty} \sqrt[k]{k!} = \lim_{k \to \infty} \frac{k}{e} = \infty$

STEP 9

Since $\lim_{k \to \infty} \frac{1}{\sqrt[k]{k!}} = 0$ , we have:
$\frac{1}{R} = 0 \implies R = \infty$
The radius of convergence of the series $\sum_{k=0}^{\infty} \frac{(x-1)^{k}}{k!}$ is $\infty$ .
Therefore, the correct answer is: D) $\infty$

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QuestionA) 0 of convergence of the series $\sum_{k=0}^{\infty} \frac{(x-1)^{k}}{k!}$ is equal to B) 1 10) The Maclaurin C) $\frac{1}{2}$ D) $\infty$

Studdy Solution

STEP 1

STEP 2

1. Identify the general term $a_k$ of the series.
2. Use the formula for the radius of convergence to find $R$ .
3. Evaluate the limit to determine the radius of convergence.

STEP 3

Identify the general term $a_k$ of the series.
$a_k = \frac{1}{k!}$

STEP 4

Apply the formula for the radius of convergence $\frac{1}{R} = \limsup_{k \to \infty} \sqrt[k]{|a_k|}$ .
$\frac{1}{R} = \limsup_{k \to \infty} \sqrt[k]{\left|\frac{1}{k!}\right|}$

STEP 5

Simplify the limit expression $\sqrt[k]{\frac{1}{k!}}$ .
$\sqrt[k]{\frac{1}{k!}} = \frac{1}{\sqrt[k]{k!}}$

STEP 6

Evaluate the limit $\lim_{k \to \infty} \sqrt[k]{k!}$ . Using Stirling's approximation $k! \approx \sqrt{2\pi k} \left(\frac{k}{e}\right)^k$ , we have:
$\sqrt[k]{k!} \approx \sqrt[k]{\sqrt{2\pi k} \left(\frac{k}{e}\right)^k}$

STEP 7

STEP 8

Combine the results to find the limit.
$\lim_{k \to \infty} \sqrt[k]{k!} = \lim_{k \to \infty} \frac{k}{e} = \infty$

STEP 9

Since $\lim_{k \to \infty} \frac{1}{\sqrt[k]{k!}} = 0$ , we have:
$\frac{1}{R} = 0 \implies R = \infty$
The radius of convergence of the series $\sum_{k=0}^{\infty} \frac{(x-1)^{k}}{k!}$ is $\infty$ .
Therefore, the correct answer is: D) $\infty$

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QuestionA) 0 of convergence of the series ∑k=0∞(x−1)kk!\sum_{k=0}^{\infty} \frac{(x-1)^{k}}{k!}∑k=0∞​k!(x−1)k​ is equal to B) 1 10) The Maclaurin C) 12\frac{1}{2}21​ D) ∞\infty∞

Studdy Solution

STEP 1

STEP 2

1. Identify the general term aka_kak​ of the series.2. Use the formula for the radius of convergence to find RRR.3. Evaluate the limit to determine the radius of convergence.

STEP 3

Identify the general term aka_kak​ of the series.ak=1k! a_k = \frac{1}{k!} ak​=k!1​

STEP 4

STEP 5

Simplify the limit expression 1k!k \sqrt[k]{\frac{1}{k!}} kk!1​​.1k!k=1k!k \sqrt[k]{\frac{1}{k!}} = \frac{1}{\sqrt[k]{k!}} kk!1​​=kk!​1​

STEP 6

STEP 7

STEP 8

Combine the results to find the limit.lim⁡k→∞k!k=lim⁡k→∞ke=∞ \lim_{k \to \infty} \sqrt[k]{k!} = \lim_{k \to \infty} \frac{k}{e} = \infty k→∞lim​kk!​=k→∞lim​ek​=∞

STEP 9

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QuestionA) 0 of convergence of the series $\sum_{k=0}^{\infty} \frac{(x-1)^{k}}{k!}$ is equal to B) 1 10) The Maclaurin C) $\frac{1}{2}$ D) $\infty$

1. Identify the general term $a_k$ of the series.
2. Use the formula for the radius of convergence to find $R$ .
3. Evaluate the limit to determine the radius of convergence.

Identify the general term $a_k$ of the series.
$a_k = \frac{1}{k!}$

Simplify the limit expression $\sqrt[k]{\frac{1}{k!}}$ .
$\sqrt[k]{\frac{1}{k!}} = \frac{1}{\sqrt[k]{k!}}$

Combine the results to find the limit.
$\lim_{k \to \infty} \sqrt[k]{k!} = \lim_{k \to \infty} \frac{k}{e} = \infty$