Why is this 'Proof' by induction not valid? 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 $S(n+1)  \lt S(n) + \frac{1}{n}$ 

 A: Other answers make the valid point 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: With induction, you can only prove $S(n)$ is true for all positive integers $n$. However, even though $S(n)$ is true for arbitrarily large $n$, the statement "$S(\infty)$" does not follow from induction because $\infty$ is not a positive integer.
A: 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 $C$.
A: 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}
A: By induction you have proved that for all $n\in\mathbb Z^+$, $\displaystyle\sum_{k=1}^n\frac 1 k$ is finite, which is true. This is not the same as proving that $\displaystyle\sum_{k=1}^\infty\frac 1 k$ is finite...
A: 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.
