$$\sum_{n=3}^{\infty}\frac{(-1)^n}{\log n}$$
Can the conditional convergence of this series be proved by alternating series test, since you need n to be a natural number for the alternating series test and here we have n is a natural number from 3 so i dont know if the condition is met
On
We can rewrite $$\sum_{n=3}^{\infty}\frac{(-1)^n}{\log n}=\sum_{n=3}^{\infty}(-1)^n \frac{1}{\log n}$$ which is in the form of $(-1)^nb_n$, which means that we can use the alternating series test. As $\frac{1}{\log n}$ never is infinity, it decreases and $$\lim_{x\to\infty}\frac{1}{\log n}=0$$ it converges.
Two points -
1) Conditional converence iff Convergence
2) For convergence of the series can we apply Alternating series test?
Well if in the series $\sum_{n=1}^{\infty}(-1)^{n}a_{n}$ if $a_{n}$ is a monotonically decreasing sequence and also $\lim_{n \rightarrow \infty} \frac{1}{\log(n)} \rightarrow 0$ so we can apply!
So is $\frac{1}{\log(n)}$ decreasing? Yes it tends to zero as $n \rightarrow \infty$,thus the series is convergent.
Well if $n$ indexing starts from $n=1$ then we have the first term $-\frac{1}{\log(1)} $ which is infinity so it diverges right!.
So if your question would have been $\sum_{n \in \Bbb{N}} (-1)^n \frac{1}{\log(n)}$ then the series would diverge!
I think you are confusing with the Indexing!
Observe that in series $\sum_{n=1}^{\infty} (-1)^{n} a_{n} = \sum_{n=k}^{\infty} (-1)^{n-k} a_{n-k}$!
So applying to your question $\sum_{n=3}^{\infty} (-1)^{n} \frac{1}{\log(n)} = \sum_{n=0}^{\infty} (-1)^{n+3} \frac{1}{\log(n+3)}$ and you can now apply the ratio test as above!