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 does not have a straightforward formula for direct evaluation, so we need to analyze its behavior as m approaches infinity.Consider (1 - 1/m)^(m^2): Let's consider the term (1 - 1/m)^(m^2) as m approaches infinity. We know that as m becomes very large, 1/m approaches 0. Therefore, the base (1 - 1/m) approaches 1.Examine Limit Resembling e: Now, we need to consider the exponent m^2. As m approaches infinity, m^2 also approaches infinity. So we have an expression of the form (1 - something small) raised to the power of something very large.Rewrite Term for Clarity: This expression resembles the limit that defines the number e, where e is the base of the natural logarithm. The limit is given by lim_(m->oo)(1 + 1/m)^m = e. However, in our case, we have a subtraction inside the parentheses, and the exponent is m^2 instead of m.Calculate Limit of (1 - 1/m)^m: We can rewrite the term as ((1 - 1/m)^m)^m to see the resemblance to the e limit more clearly. However, since we have a subtraction, the limit of (1 - 1/m)^m as m approaches infinity is not e, but 1/e. This is because (1 - 1/m)^m is the reciprocal of (1 + 1/m)^m.Evaluate (1/e)^m as m Approaches Infinity: Now, raising 1/e to the power of m, we get (1/e)^m. As m approaches infinity, (1/e)^m approaches 0 because 1/e is less than 1 and any number less than 1 raised to an infinite power tends to 0.Determine Convergence to 0: Since each term of the series (1 - 1/m)^(m^2) approaches 0 as m approaches infinity, the sum of the series is the sum of an infinite number of terms that each approach 0. This suggests that the series converges to 0.Final Result: Series Sum is 0: Therefore, the value of the infinite series sum_(m=1)^(oo)(1-(1)/(m))^(m^(2)) is 0.

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

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