Determining convergence of sum

[ag_roque]

I need to proof the convergence of:

(i is complex)
Root demonstration and Ratio demonstrations are inconclusives.

Thank you for your help.

 

[HallsofIvy]

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!

 

[Deleted member 4993]

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!

I think the problem is:

Proof of convergence of:

$\displaystyle \sum_{n=2}^\infty \frac{i^n}{log(n)}$.

where $\displaystyle i = \sqrt{-1}$

 

[HallsofIvy]

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?

 

[LCKurtz]

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?

 

[pka]

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?

Yes s/he had. Hint: it is conditionally convergent. SEE HERE

 

[ag_roque]

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?

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.

 

[pka]

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.

Think about the alternating series for the real number series. Is that applicable here?
Again look at this> look here. Seemingly it does have a sum. How can that be? Must it then converge?