Determine whether the following sequences converge and carefully justify your claims: \[(i)\quad x_n=\frac{2n\cdot n!}{n^n};\qquad (ii)\quad y_n=\sum_{k=1}^{n}\frac{\cos(k!)}{k(k+1)}.\]
Solution
Goal: Determine convergence of (i) \(x_n = \frac{2n\cdot n!}{n^n}\) and (ii) \(y_n = \sum_{k=1}^n \frac{\cos(k!)}{k(k+1)}\).
1. (i) \(x_n \to 0\).
1.1. Split the product: \(\frac{n!}{n^n} = \prod_{j=1}^{n}\frac{j}{n} \le \left(\frac12\right)^{\lfloor n/2\rfloor}\).Proof: for \(j \le n/2\), \(j/n \le 1/2\); there are at least \(\lfloor n/2 \rfloor\) such factors, and the remaining factors are \(\le 1\).
1.2. \(0 \le x_n = 2n \cdot \frac{n!}{n^n} \le 2n \left(\frac12\right)^{\lfloor n/2\rfloor} \to 0\).Proof: 1 and the squeeze theorem. Since \(\lfloor n/2\rfloor\ge (n-1)/2\), the right-hand side is at most \(2\sqrt2\,n\,2^{-n/2}\), which tends to \(0\).
1.3. Q.E.D.Proof: 2 shows \(x_n \to 0\).
2. (ii) \(y_n\) converges.
2.1. \(\left|\frac{\cos(k!)}{k(k+1)}\right| \le \frac{1}{k(k+1)}\).Proof: \(|\cos| \le 1\).
2.2. \(\sum_{k=1}^\infty \frac{1}{k(k+1)} = 1 < \infty\).Proof: telescoping: \(\frac{1}{k(k+1)} = \frac1k - \frac{1}{k+1}\), so the partial sums are \(1 - 1/(n+1) \to 1\).
2.3. \(\sum_k \frac{\cos(k!)}{k(k+1)}\) converges absolutely, hence \(y_n\) converges.Proof: comparison test with the convergent series of 2 using
2.1.
2.4. Q.E.D.Proof: 3 is convergence of \(y_n\).