91Ó°ÊÓ

To prove that the sequence is decreasing, we will need to show that \(F_{n+1} < F_{n}\) for every n. Considering the partial sums of the series: \(F_{n} = \frac{1}{1(1+n)} + \frac{1}{2(2+n)} + \frac{1}{3(3+n)} + \cdots\), and \(F_{n+1} = \frac{1}{1(1+n+1)} + \frac{1}{2(2+n+1)} + \frac{1}{3(3+n+1)} + \cdots\). Now, let's compare the individual terms of the two partial sums: \(\frac{1}{1(1+n+1)} < \frac{1}{1(1+n)}\), \( \frac{1}{2(2+n+1)} < \frac{1}{2(2+n)}\), \( \frac{1}{3(3+n+1)} < \frac{1}{3(3+n)}\), and so on. Since all the terms in the partial sum representation of \(F_{n+1}\) are less than the corresponding terms in the partial sum representation of \(F_{n}\), we can conclude that \(F_{n+1} < F_{n}\) for all n. Hence, the sequence \(\left\\{F_{n}\right\\}\) is decreasing.

To plot the sequence, we first need to calculate the first 20 components of the sequence. For these small values of n, we can obtain an approximate value of each component by calculating the sum up to a certain minimum number of terms such that the error from truncating the infinite sum is negligible. We can use a computer algebra system such as Mathematica or a powerful calculator to get the approximate values for \(F_{1}, F_{2}, \ldots, F_{20}\).

After calculating the first 20 components of the sequence, we can plot these values on a graph. You can use any graphing software, or a graphing calculator to accomplish this. The graph should show a clear decreasing trend in the sequence.

Based on the decreasing trend of the sequence and the calculated values of the components, we can make a conjecture about the limit of the sequence as n approaches infinity. Since the terms are decreasing and positive, it is reasonable to conjecture that: \(\lim _{n \rightarrow \infty} F_{n} = 0\). This conjecture will need to be proved rigorously in a more advanced mathematical setting, but based on the graphical plot, our conjecture seems reasonable for the given problem.