Evaluate sum_(m=1)((2m)!)/(m^(2m))

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Evaluate  sum_(m=1)((2m)!)/(m^(2m))

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Understand Notation and Terms: We are asked to evaluate the infinite series sum_(m=1)^∞ ((2m)!)/(m^(2m)). This is a series where each term is the factorial of twice an integer m, divided by m raised to the power of 2m.Calculate First Few Terms: The first step is to understand the notation and the terms of the series. The factorial of a number n, denoted n!, is the product of all positive integers less than or equal to n. The notation m^(2m) means m raised to the power of 2m.Evaluate Convergence: Let's calculate the first few terms of the series to get an idea of what the series looks like: For m=1: (2*1)!/(1^(2*1)) = 2!/1^2 = 2/1 = 2 For m=2: (2*2)!/(2^(2*2)) = 4!/2^4 = 24/16 = 1.5 For m=3: (2*3)!/(3^(2*3)) = 6!/3^6 = 720/729 ≈ 0.987Apply Ratio Test: We can see that as m increases, the factorial grows much faster than the exponential term in the denominator. However, the factorial is divided by a rapidly increasing exponential term, which suggests that the terms of the series decrease as m increases.Simplify Ratio Expression: To evaluate the sum, we would typically look for a pattern or a closed-form expression. However, this series does not have a simple closed-form expression, and it is not a well-known series like a geometric or arithmetic series. Therefore, we cannot find an exact sum in a simple form.Take Limit as m Approaches Infinity: Since we cannot find a closed-form expression for the sum, we can consider whether the series converges or diverges. If it converges, we can say that there is a finite sum, even if we cannot find the exact value. To test for convergence, we can use the ratio test.Conclude Convergence: The ratio test states that for a series sum_(n=1)^∞ a_n, if the limit as n approaches infinity of |a_(n+1)/a_n| is less than 1, the series converges. Let's apply the ratio test to our series: a_m = (2m)!/(m^(2m)) a_(m+1) = (2(m+1))!/((m+1)^(2(m+1))) We need to find the limit of |a_(m+1)/a_m| as m approaches infinity.Determine Sum Approximation: Calculate the ratio |a_(m+1)/a_m|: |a_(m+1)/a_m| = |((2(m+1))!/((m+1)^(2(m+1)))) / ((2m)!/(m^(2m)))| Simplify the ratio by canceling out common terms: |a_(m+1)/a_m| = |((2m+2)(2m+1)(2m)!/((m+1)^(2m+2))) / ((2m)!/(m^(2m)))| |a_(m+1)/a_m| = |((2m+2)(2m+1))/(m+1)^(2) * (m^(2m))/((m+1)^(2m))| |a_(m+1)/a_m| = |((2m+2)(2m+1))/(m+1)^(2m+2)|Now, we simplify the expression further and take the limit as m approaches infinity: |a_(m+1)/a_m| = |(4m^2 + 6m + 2)/(m+1)^(2m+2)| As m approaches infinity, the highest power of m in the numerator is m^2, and in the denominator, it is (m+1)^(2m+2), which grows much faster than m^2. Therefore, the limit of this ratio as m approaches infinity is 0.Since the limit of |a_(m+1)/a_m| as m approaches infinity is 0, which is less than 1, the ratio test tells us that the series converges.Although we have established that the series converges, we do not have a simple expression for the sum. The value of the sum would likely require numerical methods or special functions to approximate.

Evaluate $\sum_{m=1} \frac{(2m)!}{m^{2m}}$

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Evaluate $\sum_{m=1} \frac{(2m)!}{m^{2m}}$