- #1
durt
- 149
- 0
Given [itex]y_n = (2\log{2}+3\log{3}+...+n\log{n})/(n^2\log{n})[/itex], the problem is to find [itex]\lim_{n \to \infty} y_n[/itex]. I've found that
[tex]y_n \left(n-1+\frac{1}{n+1}\right) \frac{\log{n}}{\log{(n+1)}} = y_{n+1} (n+1) - 1[/tex].
So [itex] \lim_{n \to \infty} y_n(n-1) = \lim_{n \to \infty} y_{n+1} (n+1) - 1[/itex]. If I assume [itex]\lim_{n \to \infty} y_n \neq 0[/itex], separate those limits and solve to get [itex]\lim_{n \to \infty} y_n = \frac{1}{2}[/itex]. How do I show that [itex]\lim_{n \to \infty} y_n \neq 0[/itex]?
[tex]y_n \left(n-1+\frac{1}{n+1}\right) \frac{\log{n}}{\log{(n+1)}} = y_{n+1} (n+1) - 1[/tex].
So [itex] \lim_{n \to \infty} y_n(n-1) = \lim_{n \to \infty} y_{n+1} (n+1) - 1[/itex]. If I assume [itex]\lim_{n \to \infty} y_n \neq 0[/itex], separate those limits and solve to get [itex]\lim_{n \to \infty} y_n = \frac{1}{2}[/itex]. How do I show that [itex]\lim_{n \to \infty} y_n \neq 0[/itex]?