I think the problem is:Your formula did not load properly but I think it is
\(\displaystyle \sum_{n=2}^\infty \frac{i^n}{log}\).
Now, log of what? "log" is a function, not a number and has to be applied to something!
But with absolute values doesn't that go to [MATH]1[/MATH]? I think the OP indicated he had already determined that test was inconclusive, no?Thank you, that isvery plausible- though I would like ag_roque to verify that.
It seems to me like the "limit ratio test" will work: \(\displaystyle \frac{\frac{i^{n}}{log(n)}}{\frac{i^{n+1}}{log(n+1)}}\)\(\displaystyle = \frac{i^n}{i^{n+1}}\frac{log(n+1)}{log(n)}=\)\(\displaystyle -i\frac{log(n+1)}{log(n)}\). Now, does \(\displaystyle \frac{log(n+1)}{log(n)}\) converge to a number less than 1?
But with absolute values doesn't that go to [MATH]1[/MATH]? I think the OP indicated he had already determined that test was inconclusive, no?
Think about the alternating series for the real number series. Is that applicable here?yes it's inconclusive. I think log(n+1)/log(n) is equal to 1 when n tends to infinity.
that is why I am stuck. I just know how to do root and ratio demonstrations. I need another kind.