Rearranging Changes the Sum

Directly computing $T_{150}$ gives $|T_{150} - \frac{3}{2}\ln 2| \approx 0.0050 < 0.01$. (Numerical check)

Raising $n$ to $300$ shrinks the error to about $0.0025$ — the direction of convergence is consistent, not a coincidence.

The general formula is stated as known (idea level): a rearrangement repeating $p$ positive and $q$ negative terms sums to $\ln 2 + \frac{1}{2}\ln\frac{p}{q}$.

For $p = 2, q = 1$ this gives $\frac{3}{2}\ln 2$.

Its proof is not covered here.

If $\sum a_n$ converges absolutely, then for every $\varepsilon > 0$ there is an $N$ with tail $\sum_{n>N}|a_n| < \varepsilon$. (Sketch only; rigorous proof omitted)

A rearrangement's partial sums eventually contain the first $N$ terms of the original, so the two partial sums differ only by tail terms — less than $2\varepsilon$.

Since $\varepsilon$ is arbitrary, the two sums agree: when tails can be made arbitrarily small, reordering cannot change the sum.

$a_n = 1/n$ decreases to $0$. Even partial sums increase, odd ones decrease: $S_{2m+2} - S_{2m} = a_{2m+1} - a_{2m+2} > 0$ and $S_{2m+3} - S_{2m+1} = -(a_{2m+2} - a_{2m+3}) < 0$.

Now justify the sandwich.

The even sums $(S_{2m})$ increase and are bounded above by every odd sum, and the odd sums $(S_{2m+1})$ decrease and are bounded below by every even sum.

The gap is $S_{2m+1} - S_{2m} = a_{2m+1} \to 0$, so both subsequences converge to a common limit $S$ ($= \ln 2$), with $S_{2m} \le S \le S_{2m+1}$ for all $m$.

Hence for every $n$ the limit $S = \ln 2$ lies between two consecutive partial sums $S_n$ and $S_{n+1}$, so $|S - S_n| \le |S_{n+1} - S_n| = a_{n+1} = \frac{1}{n+1}$.

For $n = 100$ this gives $|S_{100} - \ln 2| \le \frac{1}{101} \approx 0.0099$ — matching the caption's figure.