Finding the sum of ln(n)/n , n=1 to infinity

Today, I will show you how to find the sum from N equals one to positive infinity of ln(n)/n. To begin, let f(x) be equal to ln(x)/x. We need to determine where f(x) starts to decrease. By taking the derivative of f(x), which is -ln(x+1)/x^2, we find that f'(x) is negative when ln(x+1)/x^2 is less than zero. Solving this inequality, we get ln(x) greater than 1, which leads to x greater than e. Since x is a discrete variable, we choose n to be 3, the smallest whole number greater than e.

Now, using the integral test, we evaluate the integral of ln(x)/x from 3 to infinity. By separating the integral as ln(x) * (1/x) dx, we let u = ln(x) and du = (1/x) dx. Substituting these values, the integral becomes ln(3) to infinity of u du. Applying the power rule to integrate, we get (1/2)u^2.

Since the upper limit is infinity, we take the limit as the upper limit approaches infinity. Plugging in ln(3) for u, we have (1/2)ln^2(3) as the final result. This tells us that the sum of ln(n)/n diverges when n is greater than or equal to 3.
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