The values $\zeta(s)$ for $s$ real should not converge to $0$ as $s\to\infty$. Is this correct? The reasoning I provide is that for $\zeta(s)$ to converge to zero, the first term of $\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}$, should be less than itself ( 1 ). But this is not possible. This is applicable only for real values of $s$.
Is this reasoning correct?
Thanks.
On
The series that you imagine defines the zeta function, does so only for arguments with real part exceeding 1; in particular, not for all real numbers, only for those exceeding 1. For other real numbers, the zeta function is not defined by the infinite series you have in mind, but rather by a process known as "analytic continuation". And when the zeta function is so defined, one finds that it does evaluate to zero at some real numbers, namely, at the negative even integers.
On
This is actually not true. Not that if $s>1$, then we have
$$\zeta(s)=\sum_{n=1}^\infty\frac1{n^s}$$
whereupon you are right that if $s>1$, then $\zeta(s)>1$. However, this is not how we define the zeta function for $s<1$. For example, at $s=0$, if we tried to use the above formula, we'd end up with
$$\zeta(0)\stackrel?=1+1+1+\dots\to\infty?$$
But the zeta function is not defined like so. Indeed, we would have $\zeta(0)=-\frac12$. One such formula for the zeta function is given by
$$\zeta(s)=\frac1{1-2^{1-s}}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}$$
And for any $0<s<1$, you would find that $\zeta(s)<0$.
Such formulas exist due to a process known as analytic continuation. If you are interested in a formula for all $s\ne1$, then
$$\zeta(s)=\frac1{1-2^{1-s}}\lim_{x\ \uparrow\ 1}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}x^{n-1}$$
And if $s$ is a negative integer, we get, due to the geometric series, that
$$\zeta(-s)=\frac1{1-2^{1+s}}\lim_{x\ \uparrow\ 1}\left[\underbrace{\frac d{dx}x\frac d{dx}x\dots\frac d{dx}x}_s\frac1{1+x}\right]$$
whereupon you will find that
$$0=\zeta(-2)=\zeta(-4)=\zeta(-6)=\dots$$
For $s>1$ the function $\zeta(s)$ is defined by
$$ \zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}.$$
When $s=1$, the series $\sum_{n=1}^\infty\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots$ is known as the harmonic series, and it diverges, so this won't work for $s=1$ or any value less than $1$. (For $s<1$, the function $\zeta(s)$ is not defined by the above series, but instead using a trick from complex analysis called analytic continuation.) You are correct that for any real number $s>1$, we have $\zeta(s)>1$. If we take the limit as $s\to+\infty$ we get $\lim\limits_{s\to\infty}\zeta(s)=1$. In fact, $\zeta(s)$ is approximately $1+1/(s-1)$.