3 ms·
That would be equal to x^y, which means that the solution is the y'th root of y? However, don't you have to prove that the series converges? Is the approach ev
by lokedhs 7y ago
That would be equal to x^y, which means that the solution is the y'th root of y?
However, don't you have to prove that the series converges? Is the approach even valid if it doesn't?
- deleted 7y ago[deleted]
- jwilk 7y agoNote that according to your solution, x=√2 for both y=2 and y=4. But of course the sequence cannot converge to both 2 and 4 at the same time!