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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.

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    That expression means the zeta function tends to infinite as $s\to1^+$. – 2017-02-02
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    Do you mind if I post a different proof of divergence to $\infty$? – 2017-02-02
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    Limit does not preserve strict inequality. What you can best hope is $+\infty \leq \lim_{\sigma \to 1^+} \zeta(\sigma) \leq +\infty$, which of course tells you that the limit is $+\infty$. – 2017-02-03
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    "But if we use the upper bound in the inequalites, the riemann zeta-function would be "bounded"" "Bounded by $+\infty$" is *not* bounded. The same argument would say that the sequence $(2^n)_n$ is bounded as $n\to\infty$, since $2^n < 3^n$ and $3^n\to \infty$. – 2017-02-03

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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

$$\zeta(\sigma)<\frac1{1-2^{1-\sigma}}\to-\infty\text{ as }\sigma\to1^-$$