Evaluate sum_(m = 1)^(oo)(1-(1)/(m))^(m^(2))=

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Evaluate  sum_(m >= 1)^(oo)(1-(1)/(m))^(m^(2))=

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Analyze Behavior as m Approaches Infinity: We are asked to evaluate the infinite series sum_(m >= 1)^(oo)(1-(1)/(m))^(m^(2)). This series is not straightforward to evaluate using standard summation techniques, but we can analyze its behavior as m approaches infinity.Consider Limit Definition of Exponential Function: Consider the term (1-(1)/(m))^(m^(2)) as m approaches infinity. We can see that (1-(1)/(m)) approaches 1 as m becomes very large. However, the exponent m^2 also becomes very large, which complicates the limit.Rewrite Term Using Exponential Function: We can rewrite the term using the limit definition of the exponential function e. Recall that (1 + x/n)^n approaches e^x as n approaches infinity. In our case, x = -1 and n = m, so we have (1 - 1/m)^m approaching e^(-1) as m approaches infinity.Consider Exponent as m Approaches Infinity: Now, we need to consider the exponent m^2. As m becomes very large, m^2 becomes even larger, and thus (1 - 1/m)^(m^2) approaches e^(-m). Since e^(-m) approaches 0 as m approaches infinity, each term in our series is approaching 0.Apply Test for Convergence: Since each term of the series approaches 0 as m approaches infinity, we might be tempted to conclude that the sum of the series is 0. However, this is not necessarily true because the series is an infinite sum, and we need to consider the rate at which the terms approach 0.Comparison Test for Convergence: To determine the behavior of the series, we can apply a test for convergence. A common test for convergence of a series involving terms with exponents is the comparison test or the ratio test. However, these tests may not be directly applicable here due to the complexity of the terms.Identify Convergent Geometric Series: Let's consider the comparison test. We need to find a series that we know converges or diverges and that our series can be compared to. A potential comparison could be the series sum_(m >= 1)^(oo) e^(-m), since e^(-m) is the limit of our terms as m approaches infinity.Use Comparison Test for Convergence: The series sum_(m >= 1)^(oo) e^(-m) is a geometric series with a common ratio less than 1 (since e^(-m) < 1 for all m > 0). Therefore, this series converges.Realize Limitations of Comparison Test: If we can show that our original series is less than or equal to the convergent geometric series term by term, then by the comparison test, our series would also converge.However, upon closer inspection, we realize that (1-(1)/(m))^(m^(2)) is not necessarily less than e^(-m) for all m. In fact, for large m, (1-(1)/(m))^(m^(2)) can be greater than e^(-m), which means we cannot use the comparison test in this way to conclude convergence.

Evaluate $\sum_{m \geq 1}^{\infty}\left(1-\frac{1}{m}\right)^{m^{2}}=$

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Evaluate $\sum_{m \geq 1}^{\infty}\left(1-\frac{1}{m}\right)^{m^{2}}=$