Why is this 'Proof' by induction not valid?

The same proof shows that the set of all positive integers is finite:

\begin{align} & \{1\} \text{ is finite.} \\ & \{1,2\} \text{ is finite.} \\ & \{1,2,3\} \text{ is finite.} \\ & \{1,2,3,4\} \text{ is finite.} \\ & \qquad \vdots \\ & \text{and so on.} \\ \text{Therefore } & \{1,2,3,4,\ldots\} \text{ is finite.} \end{align}

By induction you have proved that for all $n\in\mathbb Z^+$, $\displaystyle\sum_{k=0}^n\frac 1 k$ is finite, which is true. This is not the same as proving that $\displaystyle\sum_{k=0}^\infty\frac 1 k$ is finite...

I would add to the other comments that when you take the limit $<$ changes into $\le$. So by taking the limit you would get $\sum_{k=1}^\infty\frac{1}{k}\le\infty$, which is not particularly useful.

Other answers make the valid point made in that you can only deduce $( \forall n \in \mathbb N :P(n) )$ by induction, but not $ P(\infty) $ (though see footnote¹). There is, however, another problem in your case:

Your “$ P $” does not have the same meaning in “$ P(n)$” (where $ n\in \mathbb N $) as it does in “$ P(\infty) $”.

This is confusing, as the notation is the same, but an infinite sum is defined as a limit while a finite sum is defined inductively. Because of this, induction tells us nothing about $ P(\infty) $.

¹ _{You can sometimes deduce $P(\omega)$ when using transfinite induction, but that is a different technique and a different story.}

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 $C$.