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sum_(n=1)^(oo)((-1)^(n+1)sqrtn)/(2n-1)

77. n=1(1)n+1n2n1 \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \sqrt{n}}{2 n-1}

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Q. 77. n=1(1)n+1n2n1 \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \sqrt{n}}{2 n-1}
  1. Write Terms Pattern: The series is an alternating series, so let's write out the first few terms to see the pattern: 1123+3547+59\frac{1}{1} - \frac{\sqrt{2}}{3} + \frac{\sqrt{3}}{5} - \frac{\sqrt{4}}{7} + \frac{\sqrt{5}}{9} - \ldots
  2. Use Convergence Test: This series doesn't have a straightforward formula for summation, and it's not a geometric or arithmetic series. We need to use a convergence test to determine if it converges.
  3. Apply Alternating Series Test: Let's use the Alternating Series Test. For the series n=1(1)n+1n2n1\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\sqrt{n}}{2n-1} to converge, two conditions must be met: (11) the absolute value of the terms must be decreasing, and (22) the limit of the terms as nn approaches infinity must be zero.
  4. Check Decreasing Absolute Value: Check if the absolute value of the terms is decreasing: (1)n+1n2n1=n2n1\left|\frac{(-1)^{n+1}\sqrt{n}}{2n-1}\right| = \frac{\sqrt{n}}{2n-1}. Since the numerator grows slower than the denominator, the terms are decreasing.
  5. Check Limit as nn Approaches Infinity: Check the limit of the terms as nn approaches infinity: limn(n2n1)=0\lim_{n\to\infty}\left(\frac{\sqrt{n}}{2n-1}\right) = 0. This is true because the numerator grows at a rate of n1/2n^{1/2} while the denominator grows at a rate of nn, so the fraction approaches zero.
  6. Series Converges Without Exact Sum: Since both conditions of the Alternating Series Test are met, the series converges. However, finding the exact sum of an infinite series like this one is not straightforward and typically requires advanced calculus techniques such as power series expansion or special functions, which are beyond the scope of this problem.
  7. Series Converges Without Exact Sum: Since both conditions of the Alternating Series Test are met, the series converges. However, finding the exact sum of an infinite series like this one is not straightforward and typically requires advanced calculus techniques such as power series expansion or special functions, which are beyond the scope of this problem.Without the tools to find the exact sum, we can't provide a numerical answer. The series converges, but we don't have the sum.

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