How to write mathematical induction? Reading the literature about mathematical induction, I have learnt that there are between 4 and 3 steps in reasoning and writing the proof. I say between 3 and 4 because actually I see that texts and authors of proofs diverge slightly in their explanation and in the style in which the proof is written. This is my impression, at least.
(1) In general, we all agree that the first component is the statement we want to prove and we write it. (2) Then, it follows the base case, when we check the formula for n=1, and we see if the statement holds. So far, there is no difference between proof writers. (3) Afterwards, it follows the assumption, when we write k = n and we state that k is a generic number which appartains to N. Can be whatever natural number. (4) Finally, there is the inductive step, when we write n = k+1, and we proceed with the algebraic manipulation.
I see that some authors concentrate the steps (3) and (4) in an unique step and actually I like it more, as I don't see the need to write one passage exactly the same as the previous one, but with k in place of n! I am asking for your advice as I would like to set once for all my style in writing this kind of proof, grouping steps (3) and (4) in a unique one. However, since that I like to introduce verbally the steps, I am thinking about the right way to introduce the assumption/inductive step. Is it correct the following form, in your opinion? Assume that there exists a general number k+1, with k=n, such that... etc... I am indeed not sure about how to write this sentence, this is why I ask here.
Please, let me know if, where and why I am wrong and also you can point me out if the words I choose are the proper ones - for instance, about "general number". By the way, I am not English mother tongue speaker. 
PS. I did not use logical symbols only because I don't know how to edit them but, obviously, I would write the sentence in bold using logical connectives.
 A: Here is a case that has the step 1 in it...
THEOREM.  For $n=1,2,3,\cdots$, $$1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}<2.$$
STEP 1.  We will prove by induction: $$1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\le 2-\frac{1}{n}\tag{1}$$
STEP 2 (base case)  Since $1\le 2-\frac{1}{1}$, $(1)$ is true for $n=1$.
STEP 3 (induction hypothesis)  Now let $k \ge 1$ and assume
that $(1)$ holds for $n=k$.  That is:
$$1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{k^2}<2-\frac{1}{k}$$
STEP 4 (induction step)  Now for $n=k+1$:
$$
1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}
=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{k^2}+\frac{1}{(k+1)^2} .
$$
By the induction hypothesis, this is
$$
\le 2 - \frac{1}{k}+\frac{1}{(k+1)^2} 
< 2 - \frac{1}{k}+\frac{1}{(k+1)^2}+\frac{1}{k(k+1)}
\\
= 2-\frac{1}{k+1} = 2-\frac{1}{n} .
$$
Thus, $(1)$ holds for $n=k+1$.
STEP 5 (conclusion) Therefore, by induction, $(1)$ holds for $n=1,2,3,\dots$.
From this we get
$$
1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2} < 2.
$$
as required.
A: All the examples are great, I dont think I could contribute anything that hasn't been covered. But I thought I might add that, for me, proving the principle of induction via the well ordering principle is what really made it clear to me what was sufficient and correct when proving something via induction. 
