Is $\zeta(\sigma)\to \infty$ as $\sigma\to 1^{+}$? The series $\sum_{n=1}^{\infty}n^{-s}$ is absolutely convergent for all $\Re s>1$. Let $\zeta(s)$ be the sum. I have been taught that $\zeta(\sigma)\to \infty$ as $\sigma\to 1^{+}$. I wanted to find out why. In fact, there are the inequalites 
$$
\frac{1}{\sigma-1}<\zeta(\sigma)<\frac{1}{\sigma-1}+1
$$
which holds for all $\sigma>1$. Letting $\sigma\to 1^{+}$, we see that $\frac{1}{\sigma-1}\to \infty$. If we use the lower bound in the inequalities, we see that $\zeta(\sigma)\to \infty$ as $\sigma\to 1^{+}$. But if we use the upper bound in the inequalites, the riemann zeta-function would be "bounded" as $\sigma\to 1^{+}$. How would you explain that? Also, if we use both sides at the same time,
$$
\infty<\lim_{\sigma\to 1^{+}}\zeta(\sigma)<\infty
$$
what does this expression mean? This seems contradictory. I apologize if I ask some stupid questions, but I can't improve my understanding on my own.
 A: As a first note, the inequality
$$\infty<\lim_{\sigma\to1^+}\zeta(\sigma)<\infty$$
means that $\zeta(\sigma)\to\infty$.  One can think of this like a weird squeeze theorem.
One may, however, note that the following inequalities tell us nothing:
$$\lim_{x\to a}f(x)<+\infty$$
$$\lim_{x\to a}f(x)>-\infty$$
You are saying that it is less than infinity or greater than negative infinity.  This tells you nothing, since the limit could lie anywhere in between, including divergence to $\pm\infty$.  For example, I could tell you that
$$\lim_{x\to\infty}\sin(x)<+\infty$$
On the other hand, if I told you that
$$\lim_{x\to\infty}\sqrt{x^2-1}>+\infty$$
then this means it diverges to infinity.

Here's a different explanation of why the zeta function diverges:
$$\zeta(\sigma)=\sum_{n=1}^\infty\frac1{n^\sigma}$$
$$\eta(\sigma)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^\sigma}$$
$$\zeta(\sigma)-\eta(\sigma)=\sum_{n=1}^\infty\frac{1+(-1)^n}{n^\sigma}=\sum_{n=1}^\infty\frac2{(2n)^\sigma}=\frac2{2^\sigma}\sum_{n=1}^\infty\frac1{n^s}=2^{1-\sigma}\zeta(\sigma)$$
$$(1-2^{1-\sigma})\zeta(\sigma)=\eta(\sigma)$$
$$\zeta(\sigma)=\frac{\eta(\sigma)}{1-2^{1-\sigma}}$$
Since $\frac1{n^\sigma}\to0$ monotonically, then
$$\eta(\sigma)>1-\frac1{2^\sigma}$$
for every $\sigma>0$.  Thus,
$$\zeta(\sigma)>\frac{1-\frac1{2^\sigma}}{1-2^{1-\sigma}}\to+\infty\text{ as }\sigma\to1^+$$
You can easily modify this to show that
$$\zeta(\sigma)<\frac1{1-2^{1-\sigma}}\to-\infty\text{ as }\sigma\to1^-$$
