Inspired by Aditya Sharma

Calculus Level 4

In which step did we first make a mistake?

0 = ∑ i = 0 ∞ 0 ( 1 ) = ∑ i = 0 ∞ ( 1 − 1 ) ( 2 ) = ∑ i = 0 ∞ 1 − ∑ i = 0 ∞ 1 ( 3 ) = ( 1 + ∑ i = 1 ∞ 1 ) − ∑ j = 0 ∞ 1 ( 4 ) = 1 + ∑ j = 0 ∞ 1 − ∑ j = 0 ∞ 1 ( 5 ) = 1 + ∑ j = 0 ∞ ( 1 − 1 ) ( 6 ) = 1 ( 7 ) \begin{array} { l l l } 0 & = \displaystyle \sum_{i=0}^\infty 0 & (1) \\ & = \displaystyle \sum_{ i=0}^ \infty (1 -1) & (2) \\ & =\displaystyle \sum_{ i=0}^ \infty 1 - \displaystyle \sum_{i=0}^\infty 1 & (3) \\ & = \left ( 1 + \displaystyle \sum_{ {\color{#D61F06}i=1}}^ \infty 1\right) - \displaystyle \sum_{j=0}^\infty 1 & (4) \\ & = 1 +\displaystyle \sum_{ j=0}^ \infty 1 - \displaystyle \sum_{j=0}^\infty 1 & (5) \\ & = 1 +\displaystyle \sum_{ j=0}^ \infty (1 - 1) & (6) \\ & = 1 & (7) \\ \end{array}

Inspiration .

Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 No mistake

Calvin Lin by Calvin Lin

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1 solution

Calvin Lin Staff
Oct 27, 2016

Step 3 is NOT ok, because we are not allowed to split the sum into 2 partial sums without further justification. It leads to issues like this. In particular, we are claiming that 0 = ∞ − ∞ 0 = \infty - \infty , which is not true because the RHS is actually undefined.

Step 6 is NOT ok, because we are not allowed to combine the sum into 2 partial sums without further justification. In particular, we are claiming that ∞ − ∞ = 0 \infty - \infty = 0 . (Of course, since we want the first step, it is step 3.)

The errors result in:

0 = ( 3 ) ∞ − ∞ = 1 + ∞ − ∞ = ( 6 ) 1 0 \stackrel{(3)}{=} \infty - \infty = 1 + \infty - \infty \stackrel{(6)}{=} 1

Every other step is correct. Here are some explanations:
Step 1 - The countable sum of 0's is 0. Recall that ∑ i = 0 ∞ \sum_{i=0}^\infty is actually lim ⁡ n → ∞ ∑ i = 0 n \lim_{n\rightarrow \infty} \sum_{i=0}^n , and this finite sum is 0, so the limit is also 0.
Step 4 - We can isolate the first term of a summation. There is no reordering of terms.
Step 5 - We can relabel the indices via j = i − 1 j=i-1 and then j = i j=i in the other summation.

Note: What we are doing is taking the sequence ( a 1 − b 1 ) + ( a 2 − b 2 ) + … (a_1 - b_1) + (a_2 - b_2 ) + \ldots and reodering the terms massively into the form

a 1 + a 2 + a 3 + … − b 1 − b 2 − b 3 − … a_1 + a_2 + a_3 + \ldots - b_1 - b_2 - b_3 - \ldots

In particular, the b 1 b_1 term is moved behind infinitely many a i a_i terms. In order to justify such an arbitrary rearrangement of terms, we must ensure that the sequences are absolutely convergent . This constitutes the "proper justification" that is required.

Of course, depending on the sequence, various other "proper justification" is allowed.

10 Helpful 0 Interesting 0 Brilliant 0 Confused

Sir doesn't the first step too lead to (zero) multiplied by infinity,which itself is indeterminate,isnt that erroneous??

Chirag Shyamsundar - 4 years, 7 months ago

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The countable sum of 0's is indeed 0.

While your statement is true, it doesn't reflect the current scenario. I've edited the solution to reflect why we are not in a case of 0 × ∞ 0 \times \infty .

Calvin Lin Staff - 4 years, 7 months ago

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