Euler Maclaurin Formula and Integral Test 
Currently reading K.G.Binmore’s Mathematical Analysis. Unable to understand why the author states that this theorem shows that the infinite series and the improper integral both converge or diverge together. That is the series converges iff the improper integral converges. But the theorem says that the difference of these two converges, from that how can one say that they both converge or diverge together? 
 A: Since $\Delta_n \to 0$, for any $\epsilon > 0$ and sufficiently large $n$ we have
$$\tag{1}\int_1^n f(x) \, dx - \epsilon < \sum_{k=1}^n f(k) < \int_1^n f(x) \, dx + \epsilon,$$
$$\tag{2} \\ \sum_{k=1}^n f(k)- \epsilon < \int_1^n f(x) \, dx  < \sum_{k=1}^n f(k) + \epsilon $$
Since $f$ is positive, the integral and partial sum are increasing sequences and either converge to finite limits or diverge to $+\infty.$
If $\int_1^n f(x) \, dx \to +\infty$ then the first inequality of (1) implies that $\sum_{k=1}^n f(k) \to +\infty$. 
Similarly, if  $\sum_{k=1}^n f(k) \to +\infty$ then the first inequality of (2) implies that $\int_1^n f(x) \, dx \to +\infty$.
On the other hand, if the integral (partial sum) converges then we can apply the squeeze theorem to (1) or (2) and obtain convergence of the partial sum (integral).
For example, if $\int_1^n f(x) \, dx \to I$, then 
$$I - \epsilon < \liminf_{n \to \infty} \sum_{k=1}^n f(k) \leqslant \limsup_{n \to \infty} \sum_{k=1}^n f(k) < I + \epsilon$$
Since $\epsilon$ can be arbitrarily close to $0$, it follows that
$$\lim_{n \to \infty} \sum_{k=1}^n f(k) = I$$
