Elementary bound on the Riemann zeta function I am currently preparing for a course in analytic number theory and I wanted to  get a heads start. In my preparation, I came across the following problem:
Show that for $|y|\geq 2$, $|\zeta(1+iy)| \leq C\log|y|$ for some constant $C.$
I am very weak when it comes to determining bounds such as this, and honestly don't know where to start. This is all very new to me. Any help is appreciated.
 A: The proof I know requires more than the definition of the zeta function as a Dirichlet series. Here is an outline of the proof, along with the important estimates:
Firstly, you want to use the following defintion: $\displaystyle \zeta(s)=\frac{s}{s-1}-s\int_{1}^{\infty}\frac{\{t\}}{t^{s+1}}dt$.
Then, using Abel's summation you can show that $\displaystyle\sum_{n\leq x}\frac{1}{n^s}=\frac{s}{s-1}-\frac{1}{(s-1)x^{s-1}}-\frac{\{x\}}{x^{s}}-\int_{1}^{x}\frac{\{t\}}{t^{s+1}}dt$.
Combine the two expressions above to get $(*)$: $\displaystyle\zeta(s)= \sum_{n\leq x}\frac{1}{n^s}-s\int_{x}^{\infty}\frac{\{t\}}{t^{s+1}}dt+\frac{1}{(s-1)x^{s-1}}+\frac{\{x\}}{x^s}$.
The third fact, which is also proved using Abel's summation, is that $\displaystyle\sum_{n=1}^{x}{\frac{1}{n^s}} = O(\log(x))$ when $s=1+yi$. Indeed, show that $\displaystyle\sum_{n=1}^{x}{\frac{1}{n}} = O(\log(x))$ and then compare it to $\displaystyle\sum_{n=1}^{x}{\frac{1}{n^s}}$. If you aren't familiar with the Big O notation, this is equivalent to saying there is a $N\in\mathbb{N}$ and $c>0$ s.t. $\displaystyle \left|\sum_{n=1}^{x}{\frac{1}{n^s}}\right|\leq c\log(x)$ for all $x\geq N$.
To finish, you would need to show that the other three terms in $(*)$ don't grow faster than $\log(x)$.
A: Let $s = \sigma+ it$. For $\sigma > 1$ and by analytic continuation for $Re(s) > 0$ :
$$\zeta(s) - \sum_{n < N} n^{-s} =  s \int_N^\infty \lfloor x-N \rfloor x^{-s-1}dx = \frac{N^{1-s}}{s-1} -s \underbrace{\int_N^\infty \{x\} x^{-s-1}dx}_{\textstyle< \int_N^\infty x^{-1-\sigma}dx < N^{-\sigma}}$$
Taking $N = \lceil t \rceil$, for $\sigma > 1-\frac{A}{\log |t|}$ we have for $n \le N$ : $$n^{-\sigma} =  e^{-\sigma \ln n} < e^{-(1-\frac{A}{\log |t|}) \ln n} < e^{A} n^{-1} $$ and hence $|\sum_{n < N} n^{-s}| < \sum_{n=1}^{N-1} e^An ^{-1} <  e^{A} \ln N$

i.e. for $\sigma > 1-\frac{A}{\log |t|}$ and $t > 1$ :
$$|\zeta(s)| < 2e^A+2+\log |t|, \qquad {\scriptstyle(\text{and differentiating everything) }}\quad |\zeta'(s)| < 2e^A+2+\log^2 |t|$$

