Taylor expansion and error? This came up in  a part of the proof. 
$-\log(1-x)$ is $x$ and then want to calculate the error of this. 
The idea is that taylor series of 
$-\log(1-x)=x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+...$
We have $|x|<1$.I know how to calculate Taylor expansion, however can't see the justification from saying it is x. Next it says what is the error of this. 
Well, it has 
$x\leq \int_{1-x}^{1} \dfrac{dt}{t} \leq \dfrac{1}{1-x} x$
However, can't understand how this is true. 
This is due to trying to prove that 
$0 \leq \sum_{p\leq N} ((-log(1- \dfrac{1}{p})-\dfrac{1}{p}) \leq \sum_{p \leq N} \dfrac{1}{p(p-1)}$
 A: It is not exactly equal to $x$ except when $x=0$.  What was likely meant is that if $|x|$ is not far from $0$, then $-\log(1-x)$ is approximately equal to $x$.
One formal way of putting it is that 
$$\lim_{x\to 0} \frac{-\log(1-x)}{x}=1.$$
Take for example $x=0.1$. The calculator gives that $-\log(0.9)\approx 0.1053605$, so for $x=0.1$, $-\ln(1-x)$ is indeed fairly close to $x$.  The second term of the Taylor Series tells you roughly how big the error is when you approximate $-\log(1-x)$ by $x$. Note that if $x=0.1$, then $x^2/2=0.005$. The actual error is approximately $0.0053605$. 
Sometimes, in modelling physical situations, when we know that $|x|$ is close to $0$, we replace $-\ln(1-x)$ by $x$.  That does not mean that the two are strictly equal, only that for our purposes the approximation is good enough. 
The equation of the tangent line to $y=-\log(1-x)$ at $x=0$ turns out to be $y=x$. Recall that the tangent line to $y=f(x)$ at $x=a$ kisses the curve $y=f(x)$ at $x=a$. So if $x$ is near $0$, then $-\log(1-x)$ is very close to $x$. We call $x$ the linear approximation to $y=-\log(1-x)$ near $x=0$. Similarly, $x+\frac{x^2}{2}$ is the quadratic approximation.
Added: Your edit shows some uncertainty about the inequalities
$$x\leq \int_{1-x}^{1} \frac{dt}{t} \leq \dfrac{1}{1-x} x.$$
That uncertainty is reasonable, particularly for negative $x$. For geometric clarity we should treat the cases $x>0 and $x<0 separately.
First assume that $x$ is positive.
On our interval from $1-x$ to $1$, the function $\dfrac{1}{t}$ is always $\ge 1$. So the integral (area) is bigger than or equal to $1$ times the length of the interval, which is $x$. That proves the inequality on the left. (It is good to draw a picture.)
Similarly, on our interval, $\dfrac{1}{t}$ is always less than or equal to $\dfrac{1}{1-x}$.  So the integral is less than or equal to $\dfrac{1}{1-x}$ times the length of the interval. That yields the inequality on the right.
Next assume that $x$ is negative. It is all too easy to make mistakes with negative numbers, so we temporarily let $w=-x$. Then $w$ is positive.
We are looking at the integral from $1-x$ to $1$, so from $1+w$ to $1$.  This is is going the wrong way. But 
$$\int_{1+w}^1 \frac{dt}{t}=-\int_1^{1+w}\frac{dt}{t}.$$
By an argument close to the one given above for the case $x>0$, we find that 
$$\frac{w}{1+w}\le \int_1^{1+w}\frac{dt}{t}\le w.$$
"Multiply" the inequality by $-1$. That reverses the inequality, and we obtain
$$-w\le \int_1^{1+w}-\frac{dt}{t}\le -\frac{w}{1+w}.$$
Finally, replace $w$ by $-x$, and interchange the bounds on the integral. We get
$$x\leq \int_{1-x}^{1} \frac{dt}{t} \leq \frac{x}{1-x}.$$
A: Notice that the interval of integration in $\displaystyle\int_{1-x}^{1}\frac{\mathrm{d}t}{t}$ is $[1-x,1]$ which has a length of $x$. Note also that the integrand is between $\dfrac{1}{1-x}$ and $1$. Thus, the integral is going to be between the length of the interval times the minimum and maximum of the integrand. That is,
$$
x\le\int_{1-x}^{1}\frac{\mathrm{d}t}{t}\le\frac{x}{1-x}
$$
You could also use the Mean Value Theorem, noting that
$$
\frac{\log(1)-\log(1-x)}{x}=\frac{1}{\xi}
$$
for some $\xi$ between $1$ and $1-x$, again giving
$$
x\le-\log(1-x)\le\frac{x}{1-x}
$$
A: The inequality you give is an elementary property of the natural logarithm
$$1 - \frac{1}{x} \leqslant \log x \leqslant x - 1$$
The equality holds for $x=1$. This changing $x$ to $1-x$ gives
$$1 + \frac{1}{{x - 1}} \leqslant \log \left( {1 - x} \right) \leqslant  - x$$
And multiplying by $-1$ gives
$$\frac{x}{{1 - x}} \geqslant  - \log \left( {1 - x} \right) \geqslant x$$
which is the desired result.
Here you have an image. The blue plot is the logarithm.

You can find a proof of this first inequality in Edmund Landau's book Integral and Differential Calculus.
