Evaluate sum_(m = 1)arctan((1)/(sqrtm))=

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Evaluate  sum_(m >= 1)arctan((1)/(sqrtm))=

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Recognize Infinite Series: To evaluate the infinite sum of arctan(1/√m) for m starting from 1, we need to recognize that this is an infinite series and there is no straightforward formula to directly calculate the sum of an arctan series. However, we can check if the series converges and then possibly use numerical methods or known convergent series to estimate the sum.Check Convergence: First, we need to check if the series converges. We can use the comparison test by comparing our series to a known convergent series. The series ∑(1/√m) is a p-series with p = 1/2, which is known to diverge because p ≤ 1. However, since arctan(x) ≤ x for all x in [0,1], the series ∑arctan(1/√m) will converge if ∑(1/√m) converges. Since we know ∑(1/√m) diverges, we cannot conclude convergence from this comparison.Use Comparison Test: Another way to check for convergence is to use the integral test. The function f(m) = arctan(1/√m) is continuous, positive, and decreasing for m ≥ 1. We can set up the integral from 1 to infinity of arctan(1/√x) dx to check for convergence.Calculate Integral: We calculate the integral ∫arctan(1/√x) dx from 1 to infinity. To integrate arctan(1/√x), we can use a substitution. Let u = 1/√x, which means x = 1/u^2 and dx = -2/u^3 du. The limits of integration change accordingly: when x = 1, u = 1, and when x approaches infinity, u approaches 0.Substitution Error: Substituting into the integral, we get ∫arctan(u) * (-2/u^3) du from u = 1 to u = 0. This integral is not straightforward to evaluate, and it seems we have made a mistake in the substitution step. The correct substitution should be u = 1/√x, which means dx = -2u^3 du, not -2/u^3 du. This is a math error.

Evaluate $\sum_{m \geq 1}\arctan\left(\frac{1}{\sqrt{m}}\right)=$

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Evaluate $\sum_{m \geq 1}\arctan\left(\frac{1}{\sqrt{m}}\right)=$