$$\sum _{k=1} ^\infty \frac{(-1)^k}{k}$$
I know this question has been answered a few times but my professor has not taught alternating series test yet or anything other than ratio test, root test and comparison test where $a_i \geq 0$ for every $i \in \mathbb N$ and $\sum_{i=1} ^\infty a_i$ converges and if $|b_i| \leq a_i$ for every i then $\sum_{i=1} ^\infty b_i$ converges absolutely.
So here's my attempt using Cauchy criterion.
What we know: We say that the series $\sum _{i=1}^\infty a_i$ converges if the sequence of partial sums $(S_i)_i$$_\in$$_\mathbb N$ converges.
From Cauchy criterion, $(S_i)_i$$_\in$$_\mathbb N$ converges if and only if it is a Cauchy sequence.
It is quite obvious that $$\lim_{k\to\infty} S_k = \sum _{k=1}^\infty \frac{(-1)^k}{k}$$.
I denote $\lim_{k\to\infty} S_k = S$
Suppose ($S_k$) is convergent. Then $\forall \epsilon \gt 0, \exists N \in \mathbb N$ such that $\forall n \geq N,$
$|S_n – S|$ = $\vert \sum_{k=n+1} ^\infty \vert$ $\lt \epsilon$
I am stuck here. Is it possible to find such N for all $\epsilon \gt 0$ to hold
$\vert \sum_{k=n+1} ^\infty \vert$ $\lt \epsilon$
to be true?
(If yes, then the sequence ($S_k$) is convergent so $\sum_{k=1} ^\infty \frac{(-1)^k}{k}$ is convergent by the definition but I don't quite understand if we can always find such N)
edit: I don't think ratio test or root test are applicable to solve this and is the alternating series test the only way to solve this problem?
Best Answer
Hint. Let $$ s_n=\sum_{k=1}^n\frac{(-1)^k}{k} $$ Then observe that $$ s_1<s_3<\cdots<s_{2n-1}<s_{2n+1}<s_{2n+2}<s_{2n}<s_{2n-2}<\cdots<s_4<s_2 $$ Hence $a_n=s_{2n-1}$ is increasing and upper bounded, by $s_2$, while $b_n=s_{2n}$ is decreasing and lower bounded by $s_1$. Hence both converge, and since $a_n<b_n$, then $$ \lim a_n\le \lim b_n $$ But $b_n-a_n=\frac{1}{2n}\to 0$, and hence $$ \lim a_n= \lim b_n $$
Note. Inevitably, the idea of the proof of Alternating series test is used in the above proof.