What's wrong with this way of manipulating Grandi's series

Question

Problem: evaluate $S = 1 - 1 + 1 - 1 + 1-\cdots$

$1 = \lim_{t \rightarrow 1^{-}} t^n$ for any positive integer $n$. Here $t \rightarrow 1^{-}$ means $t \rightarrow 1$ and $t<1$.

$$S = \lim_{t \rightarrow 1^{-} } 1 - \lim_{t \rightarrow 1^{-} } t + \lim_{t \rightarrow 1^{-} } t^2 - \lim_{t \rightarrow 1^{-} } t^3 + \cdots = \lim_{t \rightarrow 1^{-}} (1-t+t^2-t^3 + t^4 - t^5 +\cdots ) = \lim_{t \rightarrow 1^{-}} \frac{1}{1+t} = \frac{1}{2}$$

What's wrong with this reasoning? It seems like the limit exists for $1 = \lim_{t \rightarrow 1^{-}} t^n$, so every step seems to follow?

Answer 1

The first time I ever saw a definition of limit of an infinite series, it said $$ \sum_{n=1}^\infty a_n = \lim_{N\to\infty} \sum_{n=1}^N a_n. $$ If it's defined that way, then $\sum_{n=1}^\infty (-1)^{n+1}$ does not exist, because the limit of the sequence of finite partial sums does not exist.

Another definition is \begin{align} & \sum_{n=1}^\infty a_n = \sum_{n\,:\,a_n\,\ge\,0} a_n + \sum_{n\,:\,a_n\,<\,0} a_n \\[10pt] = {} & \sup\left\{ \sum_{n\,\in\,A} a_n : A\text{ is a finite subset of } \{n:a_n\ge0\} \right\} \\[6pt] & {} - \sup\left\{ \sum_{n\,\in\,A} (-a_n) : A \text{ is a finite subset of } \{n:a_n<0\} \right\}. \end{align} This one makes sense only in the case where the sums of the positive and negative terms are not both infinite. In the case you consider, they are both infinite.

Thus the sum of the infinite series you are considering does not exist in either of those two senses.

One lesson to be drawn is that this example shows that $\left(\lim_t\sum_n\cdots\right)$ is not always the same as $\left(\sum_n\lim_t\cdots\right).$

When are they the same?

If $\displaystyle \sum_n \sup_t|a_n(t)| <+\infty$ then they are the same. And there are other special cases in which they are the same. Your example shows that they are not always the same.

Answer 2

It is not generally true that:

$$\lim_{t\to 1}\sum_{n=1}^{\infty} a_n(t)=\sum_{n=1}^{\infty} \lim_{t\to 1} a_n(t)$$

In general, under the standard definitions of convergence, $\sum a_n$ convergence requires $\lim_{n\to\infty} a_n=0.$

In particular,for a series to converge to $L$ requires the partial sums to get arbitrarily close to $L,$and stay close. But your series’ partial sums alternate between $0$ and $1,$ never getting close to $\frac12.$

That said, there are other definitions of infinite sums, like Cesàro summation and Ramanujan summation, where series that do not converge in the traditional sense are given values. And #\frac12$ is the value given to this series.

While these other values sometimes make sense, it is incorrect to think of it as convergence of a sum. It is more like a “principal value” for an undefined integral.

Answer 3

Since the terms you are writing form an alternating geometric series, you can consider what happens to the finite sum

$$ \sum_{k \ = \ 0}^{N} \ (-t)^k \ \ = \ \ 1 \ - \ t \ + \ t^2 \ - \ t^3 \ + \ \ldots \ + \ (-t)^N \ \ = \ \ \frac{1 \ - \ (-t)^N}{1 \ - \ (-t)} \ \ . $$

For $ \ 0 < t < 1 \ , $ this will have a directly computable value. Further, the "sub-series" of positive terms only and negative terms only are geometric series which can be summed separately and then added together to produce the result from this formula.

We find, though, that as $ \ t \ $ approaches $ \ 1 \ $ , those subseries individually sum to larger and larger positive or negative values respectively. The formula, on the other hand, gives a value $ \ \frac{1 \ - \ (-t)^N}{1 \ + \ t} \ \ $ which is finite and is the sum of the subseries sums, but oscillates through a greater "amplitude" as we increase $ \ N \ $ through successive integers on its way to giving the finite sum.

When we try to use this for $ \ t = 1 \ $ , however, this breaks down completely when we go over to the infinite sum $ \ ( N \rightarrow \infty) \ \ . $ The two subseries no longer converge, but separately go to positive or negative infinity, while the formula produces $ \ \lim_{N \ \rightarrow \ \infty} \ \frac{1 \ - \ (-1)^N}{1 \ + \ 1} \ \ . $ There was considerable controversy about how this was to be interpreted: should the sum of the infinite positive and infinite negative subseries really add up to $ \ \frac12 \ $ ? Should their sum be zero? Should they add up to any number at all? Is $ \ \lim_{N \ \rightarrow \ \infty} \ \frac{1 \ - \ (-1)^N}{1 \ + \ 1} \ $ equal to $ \frac{1 \ - \ 0}{2} \ , \ \frac{1 \ - \ 1}{2} \ , \ $ or $ \frac{1 \ + \ 1}{2} \ $ -- or what?

Even Euler at the time never settled on an answer to this; the issue of infinite series convergence was wrangled over well into the 19th-Century; Guido Grandi was writing about the series in 1703. (The way we resolve this now is to say that $ \ t = 1 \ $ is not in the interval of convergence of the infinite series, and so $ \ \frac{1}{1+t} \ $ is not represented by the series at that value of $ \ t \ . $ )