Grandi Series

The partial sums bounce between 1 and 0 forever in this scene. So this series has no sum.

Add one term and the total flips from 1 to 0, then back to 1. It never settles on one value, so it does not converge. The famous one half is only the average of the partial sums. Having an average is not the same as having a sum.

For example, pick up one 🍬 candy and put it down, over and over without end. Holding and not holding alternate forever, so you cannot say the candy is simply there.

1βˆ’1+1βˆ’1+β‹―diverges1-1+1-1+\cdots\quad\text{diverges}

Examples & Exercises

Prove directly from the partial sums that Grandi's series diverges.

Sn={1nΒ odd0nΒ evenS_n = \begin{cases}1 & n\ \mathrm{odd}\\0 & n\ \mathrm{even}\end{cases}
The partial sums bounce between 11 and 00 forever.
Show proof

  1. Sn=1βˆ’1+β‹―Β±1S_n = 1-1+\cdots\pm 1: it equals 11 for odd nn and 00 for even nn.

  2. The sequence takes only two values, alternating forever, so it has no limit.

  3. Hence the series diverges β€” no sum exists.

Prove the CesΓ ro means ΟƒN=S1+β‹―+SNN\sigma_N = \frac{S_1+\cdots+S_N}{N} converge to 12\frac{1}{2}.

ΟƒN=⌈N/2βŒ‰Nβ†’12\sigma_N = \frac{\lceil N/2\rceil}{N} \to \frac{1}{2}
Averaging settles the oscillation toward 1/21/2.
Show proof

  1. Since Sk=1S_k = 1 for odd kk and 00 for even kk, S1+β‹―+SN=⌈N/2βŒ‰S_1+\cdots+S_N = \lceil N/2\rceil.

  2. ΟƒN=⌈N/2βŒ‰N\sigma_N = \frac{\lceil N/2\rceil}{N}: for N=2mN = 2m this is m2m=12\frac{m}{2m} = \frac{1}{2}, and for N=2m+1N = 2m+1 it is m+12m+1β†’12\frac{m+1}{2m+1} \to \frac{1}{2}.

  3. Hence σN→12\sigma_N \to \frac{1}{2}. This is a mean of the partial sums — not a sum of the series.

Show that naive grouping "proves" both 00 and 11, and explain the fallacy.

(1βˆ’1)+(1βˆ’1)+β‹―=0vs1+(βˆ’1+1)+(βˆ’1+1)+β‹―=1(1-1)+(1-1)+\cdots = 0 \qquad\text{vs}\qquad 1+(-1+1)+(-1+1)+\cdots = 1
The "answer" changes with the grouping β€” both are wrong.
Show proof

  1. (1βˆ’1)+(1βˆ’1)+β‹―=0+0+β‹―=0(1-1)+(1-1)+\cdots = 0+0+\cdots = 0.

  2. But 1+(βˆ’1+1)+(βˆ’1+1)+β‹―=1+0+0+β‹―=11+(-1+1)+(-1+1)+\cdots = 1+0+0+\cdots = 1.

  3. Since both 00 and 11 "come out" of the same series, arbitrary grouping is illegitimate. Regrouping is only valid once convergence is established.

7. Series