I saw in a video this proof:

Take this equation:


and do this:

$$\begin{align} f&=1+1/2+1/4+\cdots\\ -\quad f/2&=\quad\:\:\:1/2+1/4+\cdots\\ \hline f/2&=1+0+0+\cdots \end{align}$$

This bugs me, and here's why.


We took this equation, but shifted it 1 spot over, effectively making it smaller.

Here's what I mean by shifting. Here's the same thing, but without anything shifted.

$$\begin{align} f&=\:\:\:\:1+1/2+1/4+\cdots\\ -\quad f/2&=1/2+1/4+1/8+\cdots\tag{1}\\ \hline f/2&=1/2+1/4+1/8+\cdots \end{align}$$

1: Now this part isn't shifted anymore.

Basically by shifting, I'm talking about how we align the equations. I don't like the shifting part and here's why. If we take the equation $f=1+1+1+\cdots$ and do a similar thing, first without shifting.

$$\begin{align} f&=1+1+1+\cdots\\ -\quad f&=1+1+1+\cdots\\ \hline 0&=0+0+0+\cdots \end{align}$$

Makes sense, now let's shift it.

$$\begin{align} f&=1+1+1+\cdots\\ -\quad f&=\quad\:\:\:1+1+\cdots\\ \hline 0&=1+0+0+\cdots \end{align}$$

This doesn't make any sense. $0$ doesn't equal $1$. If we were to shift it to the right a second time, we would get $0=2$.

So my question is: why is shifting allowed in that first case? Is there something I'm doing wrong in my example? Am I just not understanding infinity correctly? The way I see it, every time you shift an equation over, you lose a digit, making the answer incorrect.


4 Answers 4


We can only shift and rearrange infinite sums if both of them converge absolutely. Otherwise strange things happen, just like in your example.

  • 5
    $\begingroup$ Which points out a frequent defect in "shift the terms" proofs, because they are presented with no inkling of the fact that they rely on absolute convergence. $\endgroup$
    – David K
    Feb 27, 2015 at 15:52
  • 2
    $\begingroup$ @QxQ This is a link to some notes from an analysis course at University of Warwick that discuss the idea in some detail google.com/… I agree that it is a really weird concept at first :) $\endgroup$
    – user141592
    Feb 27, 2015 at 16:21
  • 24
    $\begingroup$ Arbitrary rearrangements do require absolute convergence to preserve the sum. But to address the OP, shifting only requires convergence. Also, adding series term-by-term only requires convergence. Moreover, some common summation methods for divergent series also respect these two operations, making them "stable" and "linear". Cesaro summation and Abel summation are good examples. So the operations are surprisingly forgiving! $\endgroup$ Feb 27, 2015 at 18:25
  • 3
    $\begingroup$ @Johanna That link blew my mind about infinite sums and enlightened me. For all those budding Buddhas out there, the example they quote is that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7} - \frac{1}{8}+\cdots=\log(2)$ (alternating positive and negative terms), but $\left(1-\frac{1}{2}\right)-\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{6}\right) - \frac{1}{8} + \left(\frac{1}{5}-\frac{1}{10}\right)-\cdots=\frac{\log(2)}{2}$ (alternating two negative for each one positive term). More generally, conditionally-convergent sums can be designed to sum to any target you want! $\endgroup$ Feb 28, 2015 at 3:32
  • 1
    $\begingroup$ @IwillnotexistIdonotexist I'm glad you liked it. It's fascinating, isn't it! $\endgroup$
    – user141592
    Feb 28, 2015 at 14:46

I feel like the answers to this question are not appropriate because they address arbitrary rearrangements of series instead of shifting. Here is an answer that is just concerned with shifting. The key lies in two facts:

Fact 1: Let $S_1 = a_1 + a_2 + \cdots$ and $S_2 = b_1 + b_2 + \cdots$ be two convergent series. Then the sum $(a_1 + b_1) + (a_2 + b_2) + \cdots$ converges and is equal to $S_1 + S_2$.

Fact 2: The series $a_1 + a_2 + \cdots$ converges if and only if the series $0 + a_1 + a_2 + \cdots$ converges.

The second fact is really just justifying the "shift" operation for a series. This proves that if you take two convergent series, you can do all of the shifting, adding (or subtracting) that you want and you will still have a valid answer. Nothing about absolute convergence is necessary for this problem.

This question about the nature of $\infty - \infty$ might be somewhat relevant.

  • $\begingroup$ Finite shifting, at least. :) $\endgroup$
    – Ian
    Feb 28, 2015 at 14:14
  • $\begingroup$ @Ian How would you shift inifinitely? $\endgroup$ Mar 1, 2015 at 12:52
  • $\begingroup$ @heinrich5991 You can't shift one term an infinite amount, but you could shift infinitely many terms each a finite amount. For instance you could replace all terms with odd index with $0$. $\endgroup$
    – Ian
    Mar 1, 2015 at 16:46

You can prove that conditionally convergent series can be rearranged to converge to any sum while any rearrangement of the terms of the absolutely convergent series must converge to the same limit that the original sequence converges to.


For any infinite sum that's convergent, shifting it doesn't change its value and for any two convergent infinite sums, the difference of their sums is equal to the sum of the differences between the terms. That's true for all convergent infinite sums, not just absolutely convergent sums. What's not true about infinite sums that are convergent but not absolutely convergent is that changing their order can't affect the sum.

Once you've proven that the sum $1 + \frac{1}{2} + \frac{1}{4}...$ is convergent, you can prove using the facts I stated earlier that shifting it over then adding the differences in the terms gives you the result of 1 and therefore the sum of the original sequence is 2. The easiest way to prove that the sum of $1 + \frac{1}{2} + \frac{1}{4}...$ is convergent is probably by first proving it equals 2 so once you've proven that, you're done and there's no need to prove again a different way that the sum is 2.

When you define $f = 1 + 1 + 1...$, you're making the wrong assumption that the infinite sum exists and that that number is called $f$. From the wrong assumption that the sum exists, you can derive the contradiction that $f - f = 0$ and $f - f = 1$.

  • $\begingroup$ "For any infinite sum that's convergent, shifting it doesn't change its value". This is the crux of the question. You may want to explain why, particularly more so since this is a 3-year-old question. $\endgroup$ Jun 21, 2018 at 16:43

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