Proving $\frac12-\frac13+\frac14-\frac15+...+\frac1{2018}-\frac1{2019} \in (\frac14, \frac13)$ What is the simplest way to prove this inequality without calculator and without calculus (I don't know calculus):
$$\frac12-\frac13+\frac14-\frac15+...+\frac1{2018}-\frac1{2019} \in \left(\frac14, \frac13\right)$$
For numbers $> \frac{1}{4}$, I can prove like this:
$$\frac12-\frac13+\frac14-\frac15+...+\frac1{2018}-\frac1{2019} > \frac12-\frac13+\frac14-\frac15+\frac16-\frac17+\frac18-\frac19=$$
$$\frac16+\frac1{20}+\frac1{42}+\frac1{72} = 0.254... > \frac{1}{4}$$
but I don't like it. Are there cleaner ways?
Edit: The original question is:
$$\frac12-\frac13+\frac14-\frac15+...+\frac1{2002}-\frac1{2003} \in \left(\frac14, \frac13\right)$$
from Challenging Problems in Algebra by Charles Salkind.
 A: First, this is (almost) a truncated Alternating Series. So the infinite result is:
$$
\sum_{n=0}^{\infty} \frac{-(-x)^n}{n} = \ln(1+x)
$$
Thus for $x=1$, we have $\ln(2)=0.693147$ as the end result. And you started at $+1/2$, 
Hence your series converges to $1-\ln(2)=0.306852$. Now, the typical epsilon-delta definition of convergence tells you that for any given $\varepsilon$ there will be $n_0$ such that the series being truncated at any term $n \gt n_0$ implies the result to be within $\varepsilon$ of the limit value. So pick $\varepsilon=0.3$.
Now, notice that THEOREM 5.14 in this reference tells us that the error in such an alternating series is no larger than the the first excluded term, which in your case is $1/2020=0.000495$. So yeah, you're well within the interval given.
A: Without using calculus, one may observe that $$S_n:=\frac12-\frac13+\frac14-\frac15\pm\ldots +\frac{(-1)^{n+1}}n $$
has the following properties:

If $n$ is even, then $S_{n+2}<S_n$. If $n$ is odd, then $S_{n+2}>S_n$.

Both follow immediately from $\frac1{n+1}-\frac1{n+2}=\frac1{(n+1)(n+2)}>0$.
By this, if $a,b<n$, $a$ odd, $b$ even, then $S_a<S_n<S_b$.
By checking, the smallest useful $a,b$ are $a=9$ (as you also found) and $b=20$ (with $S_{20}=0.331\ldots <\frac13$).
If you dislike computing $S_9$, I can only imagine what you think about computing $S_{20}$ without aid!

For a better approach, we need, still avoiding calculus, a better estimate for for $S_n-S_N$ when $N\gg n$.
We invest some manageable algebraic work in order to avoid much numerical computation.
Revisiting the above result, we have for odd $n\ge k$
$$\begin{align}S_{n+2}-S_n&=\frac1{n+1}-\frac1{n+
2}\\&=\frac1{(n+1)(n+2)}\\&\ge \frac {\frac k{k+1}}{n(n+2)}\\&=\frac k{2(k+1)}\left(\frac1n-\frac1{n+2}\right). \end{align}$$
Summing this for $n=k, k+2, k+4, \ldots$, we obtain a telescoping sum and thereby
$$ \tag1\begin{align}S_N-S_k&\ge \frac k{2(k+1)}\left(\frac1k-\frac1N\right)\\&=\frac1{2(k+1)}-\frac{k}{2(k+1)N}\end{align}$$
for $N>k$ and both odd.
The same argument leads to 
$$ \tag2S_k-S_N\ge \frac1{2(k+1)}-\frac{k}{2(k+1)N}$$
for $N>k$ and both even.
For $N=2019>9$ and $k=3$, $(1)$ leads to
$$S_N\ge \left(\frac12-\frac13\right)+\frac18- \frac 3{8N}=\frac14+\frac1{24}- \frac 3{8N}>\frac14.$$
For $N=2018>24$ and $k=4$, $(2)$ leads to 
$$S_N\le \left(\frac12-\frac13+\frac14\right)-\frac1{10}+\frac 4{10N} =\frac13-\frac1{60}+\frac 4{10N}<\frac13.$$
Therefore,
$$ \frac14<S_{2019}<S_{2018}<\frac13.$$
A: Let
$$S = \frac{1}{2}-\frac{1}{3}+\ldots +\frac{1}{2018}-\frac{1}{2019}$$
For the right side, I will use the following identity (Botez-Catalan), which can be proven with induction:
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots-\frac{1}{2n} = \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}$$
We have:
$$1-S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots-\frac{1}{2018}+\frac{1}{2019}=\frac{1}{1010}+\frac{1}{1011}+\ldots+\frac{1}{2019}$$
and using Cauchy-Schwarz:
$$
\begin{aligned}
\frac{1}{1010}+\frac{1}{1011}+\ldots+\frac{1}{2019}&\geq \frac{(1+1+\ldots+1)^2}{1010+1011+\ldots+2019}\\
&=\frac{1010^2}{505(2\cdot 2019-1009)}\\
&=\frac{2020}{3029}>\frac{2020}{3030}=\frac{2}{3}
\end{aligned}
$$
Thus, we have $S < \dfrac{1}{3}$. For the left side, I couldn't come up with a better approach than the OP.
A: For the first part, I suggest this slight modification of the proof in the OP:
$$
\frac1{2\cdot3}+\frac1{4\cdot5}+\frac1{6\cdot7}+\frac1{8\cdot9} > \frac1{2\cdot3}+\frac1{4\cdot5}+\frac1{6\cdot8}+\frac1{8\cdot10} =
\frac{40 + 12 + 5 + 3}{240} = \frac{60}{240} = \frac14,
$$
which is surely clean enough.
Next, for any $n \geqslant 2$:
\begin{align*}
\frac14 - \frac15 + \frac16 - \frac17 + \cdots + \frac1{2n} - \frac1{2n+1}
& = \frac1{4\cdot5} + \frac1{6\cdot7} + \cdots + \frac1{2n(2n+1)} \\
& < \frac1{3\cdot5} + \frac1{5\cdot7} + \cdots + \frac1{(2n-1)(2n+1)} \\
& = \frac12\left(\frac13 - \frac15 + \frac15 - \frac17 + \cdots\ + \frac1{2n-1} - \frac1{2n+1}\right) \\
& = \frac12\left(\frac13 - \frac1{2n+1}\right) \\
& < \frac16,
\end{align*}
which proves the second part.
