In a series we can take a factor that doesn't alternate and proof that it is convergent which by the definition of Leibniz criteria means the whole series is also convergent. I get that.
In lecture we shown that the non alternating factor is convergent by using the limit comparison test $\lim_{n\rightarrow \infty} \frac{c_n}{b_n} = L \in (0, \infty) \text{ and } c_n,b_n > 0 $
But what if one of the $c_n$ or $b_n$ are less than zero.
The example we have done was:
$$\sum^\infty_{n = 1}(-1)^{n-1}\tan(\frac{1}{n\sqrt{n}}) $$
where we take out $\tan(x)$ which doesn't alternate for Leibnitz criteria.
We then showed:
$$ \tan(\frac{1}{n\sqrt{n}}) = c_n \text{ is almost the same as } \frac{1}{n^{3/2}} = b_n$$
Then we solved the proof of convergence by using the limit comparison test as mentioned above.
What should we do if the other factor is negative ? Why did we use limit comparison test here instead of any other (do we always have to?) ?
Best Answer
$0<\tan\left(\frac{1}{n\sqrt n}\right)<\tan 1$ for any $n$ (positivity) and $\underset{n\to \infty }{\text{lim}} \tan\left(\frac{1}{n\sqrt n}\right)=0$ are the conditions of Leibniz test.
The comparison test is not necessary. Even $\sum _{n=2}^{\infty } (-1)^n \log n$ converges.
An alternating series must have all terms positive or all terms negative, otherwise the Leibniz test doesn't apply. If they are all negative you can collect the minus sign like the following example: $$\sum _{n=2}^{\infty } \frac{(-1)^n}{1-n^2}=-\sum _{n=2}^{\infty } \frac{(-1)^n}{n^2-1}$$