$\begin{array} { l l r } 0 & = ( 1 -1 ) & \text{ Step 1 } \\\\ & = ( 1 -1 ) + ( 1 - 1) & \text{ Step 2 } \\\\ & = ( 1 -1 ) + ( 1 - 1) + (1-1) + (1-1) + \cdots & \text{ Step 3 } \\\\ & = 1 -1 + 1 - 1 + 1 - 1 + 1 - 1 + \cdots & \text{ Step 4 } \\\\ & = 1 + (-1 + 1) + (- 1 + 1) + (-1+1) + \cdots & \text{ Step 5 } \\\\ & = 1 & \text{ Step 6 } \end{array}$

In which step above did we first introduce an error?

Step 1
Step 2
Step 3
Step 4
Step 5
Step 6
No error was made

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

The mistake was first made in Step 4, where we went from a sum with

evennumber of terms, in which the sum is indeed 0, to a sum witharbitrarynumber of terms (which has no limit of partial sums).Subsequently, in step 5, we went from a sum with an arbitrary number of terms, to a sum with

oddnumber of terms, in which the sum is indeed 1.Step 1 is justified since $0 = ( 1- 1 )$ .

Step 2 is justified since we're just adding 0.

Step 3 is justified as taking the limit, namely

$\sum_{i=1}^{\infty} (1-1) = \lim \sum_{i=1}^n (1-1 ) = \lim 0 = 0$

Step 4 is wrong as explained above.

Step 5 is wrong as explained above.

Step 6 is justified.