Grandi Series

The partial sums bounce between 1 and 0 forever, so this series has no sum. 12\frac{1}{2} is only the average of the partial sums (the Cesàro mean), not a sum of the series.

1−1+1−1+⋯diverges1-1+1-1+\cdots\quad\text{diverges}

Examples & Exercises

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

Sn={1n odd0n 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=1−1+⋯±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 — no sum exists.

Prove the Cesàro means σN=S1+⋯+SNN\sigma_N = \frac{S_1+\cdots+S_N}{N} converge to 12\frac{1}{2}.

σN=⌈N/2⌉N→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+⋯+SN=⌈N/2⌉S_1+\cdots+S_N = \lceil N/2\rceil.
  2. σN=⌈N/2⌉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+1→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.

(1−1)+(1−1)+⋯=0vs1+(−1+1)+(−1+1)+⋯=1(1-1)+(1-1)+\cdots = 0 \qquad\text{vs}\qquad 1+(-1+1)+(-1+1)+\cdots = 1
The "answer" changes with the grouping — both are wrong.
Show proof
  1. (1−1)+(1−1)+⋯=0+0+⋯=0(1-1)+(1-1)+\cdots = 0+0+\cdots = 0.
  2. But 1+(−1+1)+(−1+1)+⋯=1+0+0+⋯=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.

6. Series