Final answer:
The integral of xe^(-x^2) from 1 to infinity is evaluated using substitution, which results in a convergent improper integral. The sum ke^(-k^2) from n equals 1 to infinity can be shown to be convergent using the comparison test because the terms of the series approach zero as n increases.
Step-by-step explanation:
The question requires evaluating an improper integral and determining the convergence or divergence of an infinite series. To find the value of ∫ from 1 to infinity of xe⁻²dx, we need to use integration by parts or substitution to evaluate the improper integral. Concerning the sum from n equals 1 to infinity of (ke⁻k²), we will employ a convergence test such as the comparison test or ratio test to ascertain whether the series is convergent or divergent.
For the integral, we can let u = x² which gives du = 2xdx, and the integral transforms into ½∫ from 1 to infinity of e⁻udu, which converges as it represents the tail of the exponential distribution. For the infinite series, if k is a constant, the terms approach zero due to the exponential function as n increases, and we compare it to an exponential decay that is known to be convergent.