Let's solve the problem **step by step** using the **Integral Test**. --- ### Given Series \[ \sum_{n=1}^{\infty} n^k e^{-n} \] where \( k \) is a positive integer. --- ## **Step 1: Define the Function** Let \( f(x) = x^k e^{-x} \) for \( x \geq 1 \). - \( f(x) \) is **positive, continuous, and decreasing** for \( x \geq 1 \) (for all \( k > 0 \)). --- ## **Step 2: Set Up the Integral Test** The Integral Test says: If \( \int_1^{\infty} f(x) \, dx \) converges, the series converges. If the integral diverges, the series diverges. So evaluate: \[ \int_1^{\infty} x^k e^{-x} \, dx \] --- ## **Step 3: Analyze the Integral** This is a classic improper integral. For \( k \) a positive integer, this resembles the **Gamma function**: \[ \Gamma(s) = \int_0^{\infty} x^{s-1} e^{-x} dx \] But our lower limit is 1, not 0. Let's relate it to the Gamma function: \[ \int_1^{\infty} x^k e^{-x} dx = \int_0^{\infty} x^k e^{-x} dx - \int_0^1 x^k e^{-x} dx \] - \(\int_0^{\infty} x^k e^{-x} dx = \Gamma(k+1) = k!\) (finite for positive integers!) - \(\int_0^1 x^k e^{-x} dx\) is finite for all \( k \). So, \(\int_1^{\infty} x^k e^{-x} dx\) is **finite** for any positive integer \( k \). --- ## **Step 4: Conclusion by the Integral Test** Since the improper integral converges, **the series converges**. --- ## **Final Answer** \[ \boxed{ \sum_{n=1}^{\infty} n^k e^{-n} \text{ converges for any positive integer } k. } \] --- **Summary of Steps:** 1. Set up \( f(x) = x^k e^{-x} \). 2. Apply the Integral Test and relate the integral to the Gamma function. 3. The integral converges, so the series converges.

Let's go through a **detailed step-by-step solution** to determine whether the series \[ \sum_{n=1}^{\infty} n^k e^{-n} \] converges or diverges, using the **Integral Test**. --- ### Step 1: Define the function \(f(x)\) For \(x \geq 1\), define: \[ f(x) = x^k e^{-x} \] where \(k\) is a positive integer. **Note:** - \(f(x)\) is positive for \(x \geq 1\), - \(f(x)\) is continuous, - \(f(x)\) is decreasing for sufficiently large \(x\) because exponential decay dominates polynomial growth. --- ### Step 2: Verify conditions for the Integral Test - Since \(f(x)\) is positive, continuous, and decreasing for \(x \geq 1\), the Integral Test applies. --- ### Step 3: Set up the improper integral Evaluate: \[ I = \int_1^{\infty} x^k e^{-x} \, dx \] - The nature of this integral determines the convergence of the series. --- ### Step 4: Recognize the integral as a Gamma function Recall the **Gamma function**: \[ \Gamma(s) = \int_0^{\infty} x^{s-1} e^{-x} \, dx \] - For positive integers \(k\), \(\Gamma(k+1) = k!\). Note that: \[ I = \int_1^{\infty} x^k e^{-x} \, dx \] which relates to the Gamma function: \[ \int_0^{\infty} x^k e^{-x} \, dx = \Gamma(k+1) = k! \] Thus, \[ I = \int_1^{\infty} x^k e^{-x} \, dx = \left(\int_0^{\infty} x^k e^{-x} \, dx \right) - \left(\int_0^{1} x^k e^{-x} \, dx \right) = k! - \int_0^{1} x^k e^{-x} \, dx \] --- ### Step 5: Evaluate the remaining integral over \([0, 1]\) \[ J = \int_0^1 x^k e^{-x} \, dx \] - Since \(x \in [0,1]\), \(e^{-x}\) is bounded between \(e^{-1}\) and 1. - The integrand \(x^k e^{-x}\) is continuous and finite over \([0,1]\). Therefore, \(J\) is finite. - In fact, \(J\) can be expressed as a convergent integral and is finite. --- ### Step 6: Conclude the convergence of the integral Since: \[ I = k! - J \] and both \(k!\) and \(J\) are finite, it follows that: \[ I = \int_1^{\infty} x^k e^{-x} \, dx \] is **finite**. **Hence, the integral converges**. --- ### **Step 7: Apply the Integral Test conclusion** Because: \[ \int_1^{\infty} f(x) \, dx \text{ converges} \] the **series** \[ \sum_{n=1}^{\infty} n^k e^{-n} \] **also converges**. --- ### **Final Answer:** \[ \boxed{ \text{The series } \sum_{n=1}^{\infty} n^k e^{-n} \text{ converges for any positive integer } k. } \] --- ### **Summary of reasoning:** - The function \(f(x) = x^k e^{-x}\) meets the criteria for the Integral Test. - The integral \(\int_1^{\infty} x^k e^{-x} dx\) is finite because it relates to the Gamma function. - Therefore, the series converges for all positive integers \(k\). --- **If you need any further clarification or detailed calculations of the integral, feel free to ask!**