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Evaluate m1(11m)m2=\sum_{m \geq 1}^{\infty}\left(1-\frac{1}{m}\right)^{m^{2}}=

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Q. Evaluate m1(11m)m2=\sum_{m \geq 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 m1(11m)m2\sum_{m \geq 1}^{\infty}(1-\frac{1}{m})^{m^{2}}. This series is not straightforward to evaluate using standard summation techniques, but we can analyze its behavior as mm approaches infinity.
  2. Consider Limit Definition of Exponential Function: Consider the term (11m)m2(1-\frac{1}{m})^{m^{2}} as mm approaches infinity. We can see that (11m)(1-\frac{1}{m}) approaches 11 as mm becomes very large. However, the exponent m2m^2 also becomes very large, which complicates the limit.
  3. Rewrite Term Using Exponential Function: We can rewrite the term using the limit definition of the exponential function ee. Recall that (1+xn)n(1 + \frac{x}{n})^n approaches exe^x as nn approaches infinity. In our case, x=1x = -1 and n=mn = m, so we have (11m)m(1 - \frac{1}{m})^m approaching e1e^{-1} as mm approaches infinity.
  4. Consider Exponent as mm Approaches Infinity: Now, we need to consider the exponent m2m^2. As mm becomes very large, m2m^2 becomes even larger, and thus (11/m)m2(1 - 1/m)^{m^2} approaches eme^{-m}. Since eme^{-m} approaches 00 as mm approaches infinity, each term in our series is approaching 00.
  5. Apply Test for Convergence: Since each term of the series approaches 00 as mm approaches infinity, we might be tempted to conclude that the sum of the series is 00. 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 00.
  6. 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.
  7. 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 m1em\sum_{m \geq 1}^{\infty} e^{-m}, since eme^{-m} is the limit of our terms as mm approaches infinity.
  8. Use Comparison Test for Convergence: The series m1em\sum_{m \geq 1}^{\infty} e^{-m} is a geometric series with a common ratio less than 11 (since em<1e^{-m} < 1 for all m>0m > 0). Therefore, this series converges.
  9. 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.
  10. 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))m2(1-(1)/(m))^{m^{2}} is not necessarily less than eme^{-m} for all mm. In fact, for large mm, (1(1)/(m))m2(1-(1)/(m))^{m^{2}} can be greater than eme^{-m}, which means we cannot use the comparison test in this way to conclude convergence.

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