What is the error in this fake proof which uses series to show that $1=0$? A common "trick" for obtaining a closed form of a geometric series is to define
$$ R := \sum_{k=0}^{\infty} r^k, $$
then manipulate the series as follows:
\begin{align}
R - rR
&amp;= \sum_{k=0}^{\infty} r^{k} + \sum_{k=0}^{\infty} r^{k+1} \\
&amp;= (1 + r + r^2 + r^3 + \dotsb) - (r + r^2 + r^3 + \dotsb) \\
&amp;= 1 + (r + r^2 + r^3 + \dotsb) - (r + r^2 + r^3 + \dotsb) \\
&amp;= 1.
\end{align}
On the other hand, $R-rR = (1-r)R$.  Hence
$$ (1-r)R = 1
\implies R = \frac{1}{1-r}. $$
In this example, the formula is obtained by a sequence of relatively elementary algebraic manipulations.

By a similar kind of manipulation, suppose that
$$ S := 1 + 1 + 1 + 1 + \dotsb = \sum_{k=0}^{\infty} 1. $$
$S$ is unaffected by addition of $1$, and so $S = 1+S$.  Canceling $S$ from both sides gives $0 = 1$, which is clearly nonsense.


  Question: What went wrong with the second computation?  Why do these arguments work well for summing the geometric series, but not for the series of ones?
Your argument hinges upon the assumption that $S$ is a number you can do arithmetic with. It isn't a number, you can't do arithmetic with it, and this is what you have shown (by contradiction).
In fact, for $|r|&lt;1$,
$$R=\lim_{n\to\infty}\sum_{k=0}^n r^k$$ and
$$(1-r)R=\lim_{n\to\infty}\left(\sum_{k=0}^n r^k-\sum_{k=0}^n r^{k+1}\right)=\lim_{n\to\infty}(1-r^{n+1})=1.$$ All steps are rigorous.

If we try a similar argument with
$$S=\lim_{n\to\infty}\sum_{k=0}^n1$$
we obtain
$$S+1=\lim_{n\to\infty}\sum_{k=0}^{n+1}1$$
and by substraction,
$$S+1-S=\lim_{n\to\infty}\left(\sum_{k=0}^{n+1}1-\sum_{k=0}^n1\right)=\lim_{n\to\infty} 1\ne0.$$