I'm doing some practice problems for alternating series test. I understand where it uses the root test, but not where they use L'H rule and henceforth. Why is x = 1/ n and how was L'H rule applied here? Would anyone kindly break it down or even provide a better/simpler way of solving this series?
On
L’Hospital rule is applied to the ratio
$$\frac{\log(1-x)}{x}$$
by $(\log(1-x))’=\frac{-1}{1-x}$.
Root test in this case is fine, as an alternative way since
$$|a_n|\sim \frac1{e^{n}}$$
we can use limit comparison test with $\sum \frac1{n^2}$.
On
This set is Absolute convergent. Because:
Let your series be $(s_n)=(-1^n)(1-\dfrac{1}{n})^{n^2}=(-1^n)(a_n)=(-1^n)e^{n^2\ln(1-\frac{1}{n})}$ $$n=1\implies(a_1)=0$$ So we only have to show the sum from $n=2$ to $\infty$ converges. $$|a_2|=(\dfrac{1}{2})^{4}>(\dfrac{2}{3})^{9}=|a_3|$$ By using mathematical induction, you can prove the following statement holds: $$|a_2|>|a_3|>...>|a_n|\ge0$$Which means $(a_n)$ is monotonic and bounded below by $0$. And:$$n^2\ln(1-\frac{1}{n})\to -\infty, \forall n\to\infty$$ $$\implies e^{n^2\ln(1-\frac{1}{n})}\to0$$ Hence, it converges to $0$.
Since $e^{n^2\ln(1-\frac{1}{n})}<1,\forall n>1$.So By using Comparison test (compare $(a_n)$ to $\frac{1}{n^2}$), we can prove that $$a_n\le\dfrac{1}{n^2},\forall n\ge1$$ Since $\sum_{n=1}^{\infty}\dfrac{1}{n^2}$ converges, $\sum_{n=1}^{\infty}(a_n)$ also converges.
Since $(s_n)$ has a term $(-1^n)$, This series is absolutely convergent.
And I want to explain how the author uses L'H rule for you:
Since we can't easily get the limit of $n\ln(1-\frac{1}{n})$, we consider to use L'h rule to solve it. Therefore we need a fraction to reinterpret that expression. but that expression contains the term $\frac{1}{n}$ so we use $x$ to represent it since $n\to\infty$ is the same as $x\to0$. Did I solve your problem?
$\lim_{x \to 0} \frac {\log (1-x)} x =\lim_{x\to 0} \frac {-1/(1-x)} {1}=-1$ by L'Hopital's Rule.
This implies that $\lim_{n \to \infty} \frac {\log (1-\frac 1 n)} {\frac 1 n} =-1$.