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Evaluate m=1(11m)m2\sum_{m=1}^{\infty}\left(1-\frac{1}{m}\right)^{m^{2}}

Full solution

Q. Evaluate m=1(11m)m2\sum_{m=1}^{\infty}\left(1-\frac{1}{m}\right)^{m^{2}}
  1. Analyze Behavior as mm Approaches Infinity: We are asked to evaluate the infinite series m=1(11m)m2\sum_{m=1}^{\infty}(1-\frac{1}{m})^{m^{2}}. This series does not have a straightforward formula for direct evaluation, so we need to analyze its behavior as mm approaches infinity.
  2. Consider (11/m)m2(1 - 1/m)^{m^2}: Let's consider the term (11/m)m2(1 - 1/m)^{m^2} as mm approaches infinity. We know that as mm becomes very large, 1/m1/m approaches 00. Therefore, the base (11/m)(1 - 1/m) approaches 11.
  3. Examine Limit Resembling ee: Now, we need to consider the exponent m2m^2. As mm approaches infinity, m2m^2 also approaches infinity. So we have an expression of the form (1something small)(1 - \text{something small}) raised to the power of something very large\text{something very large}.
  4. Rewrite Term for Clarity: This expression resembles the limit that defines the number ee, where ee is the base of the natural logarithm. The limit is given by limm(1+1m)m=e\lim_{m\to\infty}(1 + \frac{1}{m})^m = e. However, in our case, we have a subtraction inside the parentheses, and the exponent is m2m^2 instead of mm.
  5. Calculate Limit of (11/m)m(1 - 1/m)^m: We can rewrite the term as ((11/m)m)m((1 - 1/m)^m)^m to see the resemblance to the ee limit more clearly. However, since we have a subtraction, the limit of (11/m)m(1 - 1/m)^m as mm approaches infinity is not ee, but 1/e1/e. This is because (11/m)m(1 - 1/m)^m is the reciprocal of (1+1/m)m(1 + 1/m)^m.
  6. Evaluate (1/e)m(1/e)^m as mm Approaches Infinity: Now, raising 1/e1/e to the power of mm, we get (1/e)m(1/e)^m. As mm approaches infinity, (1/e)m(1/e)^m approaches 00 because 1/e1/e is less than 11 and any number less than 11 raised to an infinite power tends to 00.
  7. Determine Convergence to 00: Since each term of the series (11m)m2(1 - \frac{1}{m})^{m^2} approaches 00 as mm approaches infinity, the sum of the series is the sum of an infinite number of terms that each approach 00. This suggests that the series converges to 00.
  8. Final Result: Series Sum is 00: Therefore, the value of the infinite series m=1(11m)m2\sum_{m=1}^{\infty}\left(1-\frac{1}{m}\right)^{m^{2}} is 00.

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