TITLE: Why is this 'Proof' by induction not valid?
QUESTION [39 upvotes]: I am trying to understand why induction is valid. For instance why would this 'proof' not be valid under the principle of proof by induction ? :

$$ \sum_{k=1}^{\infty} \frac{1}{k} \lt \infty$$ because using induction on the statement 
$$S(n) =  \sum_{1}^{n} \frac{1}{k} \lt \infty$$
- "$S(1) < \infty$ is true and "$S(n) < \infty$" implies "$S(n+1) < \infty$" since $S(n+1)  \lt S(n) + \frac{1}{n}$

REPLY [5 votes]: A lot of people are talking about the meaning of $P(\infty)$, which I believe is something of a red herring: usually the symbol $\infty$ is a notational convenience, and has no meaning as a formal object.
In particular:

$\sum_{n=0}^\infty x_n$ means $\lim_{m\rightarrow\infty}\sum_{n=0}^m x_n$, where again, the limit towards $\infty$ has a precise mathematical meaning not involving the symbol $\infty$.
$x<\infty$ simply means that there is some $C\in\mathbb{R}$ such that $x<C$. Alternately, for an increasing sequence $x_n$, $\lim_{n\rightarrow\infty}x_n<\infty$ means that there is a $C\in\mathbb{R}$ such that $x_n<C$ for every $n$.

Now if you try to prove that there is some $C$ such that for every $m$, $\sum_{n=0}^m \frac{1}{n}<C$, you'll run into trouble: certainly there is some such $C$ for each $m$, but there is no $C$ that works uniformly for every $m$. Indeed, if $x<C$, then $x+\frac{1}{m}$ could very well be above $C$.