Proving a series is convergent - $\sum _{n=1} ^\infty \frac{(-1)^n}{n}$ without using alternating series test $$\sum _{k=1} ^\infty \frac{(-1)^k}{k}$$
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?
 A: 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.
A: 
Not only can we show that the series of interest converges, we can also evaluate it using elementary tools only. We now present two approaches.


METHODOLOGY $1$:
Noting that $\int_0^1 x^{n-1}\,dx=\frac1n$, we can write
$$\begin{align}
\sum_{n=1}^N \frac{(-1)^n}{n}&=\sum_{n=1}^N (-1)^n\int_0^1 x^{n-1}\,dx\\\\
&=-\int_0^1 \sum_{n=1}^N (-x)^{n-1}\,dx\\\\
&=-\int_0^1 \frac{1-(-x)^N}{1+x}\,dx\\\\
&=-\log(2)+(-1)^N\int_0^1 \frac{x^N}{1+x}\,dx
\end{align}$$
Since $\left|\frac{x^N}{1+x}\right|\le x^N$, we have the estimate
$$\left|\int_0^1 \frac{x^N}{1+x}\,dx\right|\le \frac{1}{N+1}$$
Therefore, 
$$\lim_{N\to\infty }\sum_{n=1}^N \frac{(-1)^n}{n}=-\log(2)$$

METHODOLOGY $2$:
$$\begin{align}
\sum_{n=1}^{2N}\frac{(-1)^n}{n}&=\sum_{n=1}^N \frac1{2n}-\sum_{n=1}^N \frac1{2n-1}\\\\
&=\sum_{n=1}^N \frac1{2n}-\left(\sum_{n=1}^{2N}\frac1n-\sum_{n=1}^N\frac1{2n}\right)\\\\
&=-\sum_{n=N+1}^{2N}\frac1n\\\\
&=-\sum_{n=1}^N \frac{1}{n+N}\\\\
&=-\frac1N \sum_{n=1}^N\frac1{1+(n/N)}
\end{align}$$
The last expression is the Riemann Sum for $-\int_0^1 \frac1{1+x}\,dx=-\log(2)$ as expected.

If these methodologies are not quite the way forward that the OP is seeking, then we simply note 
$$\begin{align}
\left|\sum_{n=1}^{2N}\frac{(-1)^n}{n}\right|&=\left|\sum_{n=1}^N \frac1{2n}-\sum_{n=1}^N \frac1{2n-1}\right|\\\\
&=\sum_{n=1}^N \frac{1}{2n(2n-1)}\\\\
&\le \frac12 \sum_{n=1}^N \frac1{n^2}
\end{align}$$
and the series of interest converges by comparison with the series $\sum_{n=1}^\infty \frac{1}{n^2}$ (Note that $\sum_{n=2}^\infty \frac{1}{n^2}\le \sum_{n=2}^\infty \frac{1}{n(n-1)}=1$).
A: We can process the terms in pairs because two successive partial sums differ in $\pm\dfrac1n$, which tends to zero.
Now let us study the series with general term
$$\frac1{2n}-\frac1{2n+1}=\frac1{(2n+1)2n}\le\frac1{4n^2}.$$
For this upper bound, we will group the second and third terms, then the fourth to seventh, and so on, each time doubling the length.
$$4S=1+\left(\frac1{2^2}+\frac1{3^2}\right)+\left(\frac1{4^2}+\frac1{5^2}+\frac1{6^2}+\frac1{7^2}\right)+\cdots
\\\le1+\left(\frac1{2^2}+\frac1{2^2}\right)+\left(\frac1{4^2}+\frac1{4^2}+\frac1{4^2}+\frac1{4^2}\right)+\cdots
\\=1+\frac12+\frac14+\cdots,$$ a well known convergent series.
