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I have learned that positive infinity plus negative infinity isn't equal to zero, it's an indeterminate form. However what happens if we subtract two infinite divergent series $\displaystyle{\sum_{n=1}^{\infty} n=\infty}$ ? Is it sill an indeterminate form or is it zero?

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

Davide Giraudo
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achichi
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5 Answers5

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It's still indeterminate. The moment one piece of an expression diverges, the entire expression diverges.

In the equation you wrote, the equality $$\sum_{n=1}^\infty n - \sum_{n=1}^\infty n = \sum_{n=1}^\infty (n-n)$$ is not valid. This is because in order for it to work, one must rearrange infinitely many terms and the infinite commutative law is simply false without further hypothesis.

That is, you know that $a+b = b + a$ for any two real numbers $a$ and $b$, and this can be extended to say that the sum of a finite number of real numbers is independent of the order you sum them up. It does not follow (nor is it true) that infinite sums can be rearranged and keep the same sum.

See http://en.wikipedia.org/wiki/Absolutely_convergent#Rearrangements_and_unconditional_convergence for a precise formulation of when the infinite commutative law is valid and see http://en.wikipedia.org/wiki/Riemann_series_theorem for information on how it fails in general.

Jason DeVito - on hiatus
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You should always be worried when working with infinite quantities, and by "worried", I mean that you should be extra careful.

In your particular case, the quantity $\left(\sum_{n=1}^\infty n\right) - \left(\sum_{n=1}^\infty n\right)$ is an indeterminate, because it does not make sense to give it any particular value. The problem is that the value of your expression changes if we rearrange the terms. Notice that "$\sum_{n=1}^\infty n$" is a symbol, that stands for the limit of the partial sums. As such, when you write:

$$\left(\sum_{n=1}^\infty n\right) - \left(\sum_{n=1}^\infty n\right)= \sum_{n=1}^\infty (n - n) = 0$$ you have implicitly made an arbitrary choice on how you are adding the corresponding partial sums. But we could make another choice so that:

$$\left(\sum_{n=1}^\infty n\right) - \left(\sum_{n=1}^\infty n\right) = (1+2 - 1)+(3+4 - 2) + (5+6 -3) + \cdots $$ $$=\sum_{k=1}^\infty ((2k-1)+2k-k) = \sum_{k=1}^\infty (3k-1) = \infty$$

which clearly diverges to infinity. Or we could have chosen another way:

$$\left(\sum_{n=1}^\infty n\right) - \left(\sum_{n=1}^\infty n\right) = (1 - 1-2)+(2 - 3-4) + (3 -5-6) + \cdots $$ $$=\sum_{k=1}^\infty (k-(2k-1)-2k) = \sum_{k=1}^\infty -(3k-1) = -\infty.$$

Since we cannot decide which one of the three possibilities is right (all three are simply arbitrary rearrangements of partial sums), we cannot give this quantity a determined value, and hence we call it an indeterminate.

Álvaro Lozano-Robledo
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  • Thnx, I would give it +1, if I could. – achichi Aug 24 '11 at 19:40
  • There are difficulties with this argument. For look at $\left(\sum_0^\infty \frac{(-1)^n}{n}\right)-\left(\sum_0^\infty \frac{(-1)^n}{n}\right)$. This is $0$, but rearrangements can yield anything, since the series are not absolutely convergent. – André Nicolas Aug 24 '11 at 21:33
  • @André Nicolas, in my answer I am simply pointing out that there is a problem with writing $\left(\sum_{n=1}^\infty n\right) - \left(\sum_{n=1}^\infty n\right)= \sum_{n=1}^\infty (n - n)$, because the equal sign is tantamount to a rearrangement of the series. Then I explain how different rearrangements yield different solutions. I agree $(\sum_{n\geq 1} (-1)^n/n) - (\sum_{n\geq 1} (-1)^n/n)=0$, but if you were going to say $(\sum_{n\geq 1} (-1)^{n}/n) - (\sum_{n\geq 1} (-1)^n/n)=\sum_{n\geq 1} (\text{some rearrangement})$, then my comments above would apply once again. – Álvaro Lozano-Robledo Aug 25 '11 at 00:43
  • (And note, by the way, that what you wrote is $\infty-\infty$ and not zero, and not what you meant to write, i.e., $n=1$ to $\infty$.) – Álvaro Lozano-Robledo Aug 25 '11 at 00:47
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Consider different grouping of terms:

$$ (1-1) + (2-2) + (3-3) + \ldots = 0 + 0 + \ldots = 0 $$

$$ 1 + (2-1) + (3-2) + \ldots = 1 + 1 + 1 + \ldots = + \infty $$

Sasha
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In some cases term-by-term subtraction of series that diverge to $\infty$ yields an absolutely convergent series. A well-known example is this: $$ \sum_{n=1}^\infty \frac1n $$ $$ 0+\sum_{n=2}^\infty \left(\log_e\frac{n}{n-1}\right) $$ (I prepended "$0+{}$" to make clear which is the $N$th term in each series for purposes of term-by-term subtraction.) The term-by-term difference converges absolutely to the Euler--Mascheroni constant.

Michael Hardy
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Divergent Series by GH Hardy is a classic text on what can be done to assign a consistent sum to divergent (conditionally convergent) series (for example in Fourier Analysis). This is probably more sophisticated than the question requires, but for further exploration of an interesting area, well worth consulting.

Mark Bennet
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