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

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

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Analyze Series Behavior: We are asked to evaluate the infinite series sum_(m >= 1)(1-(1)/(m))^(m^(2)). This series is not a standard series with a known closed form, so we need to analyze its behavior as m increases.Consider Large Terms: Let's consider the term (1-(1)/(m))^(m^(2)) as m becomes very large. The base (1-(1)/(m)) approaches 1 as m goes to infinity, and the exponent m^2 increases much faster than m. This suggests that each term in the series is of the form (1 - something small) raised to the power of (something very large).Use Limit Definition of e: We can use 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 we can rewrite our term as ((1 - 1/m)^m)^m. As m approaches infinity, (1 - 1/m)^m approaches e^(-1).Evaluate Exponential Term: However, we have an additional m power to consider. So, our term becomes (e^(-1))^m = e^(-m). As m becomes large, e^(-m) approaches 0 because the exponential function decays rapidly.Conclude Series Convergence: Since each term of the series approaches 0 as m increases, and the terms are positive and decreasing, we can conclude that the series converges. The sum of the series is the limit of the partial sums as m approaches infinity.Determine Sum as 0: The limit of e^(-m) as m approaches infinity is 0. Therefore, the sum of the series is the sum of an infinite number of terms that each approach 0. This implies that the sum of the series is 0.

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

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