10.6.5 Using Geometric Series to Estimate Tails:
           Logarithmic Series

Here again is the Taylor series for `ln text[(] 1+x text[)]`:

x - x 2 2 + x 3 3 - x 4 4 + ⋅⋅⋅ .

From Activity 2 we know that this power series converges if `0<x<=1` and diverges if `x>1` or `x<-1`. For `x=-1`, this series is the negative of the harmonic series, which also diverges. But we don't know for sure whether it converges for `-1<x<0`. Our technique of comparing tails with geometric series will resolve this question also.

For `x` in the interval `-1<x<0`, let's write `t=-x= text[|] x text[|]`. Then `0<t<1`. When we substitute `x=-t` in the power series for `ln text[(] 1+x text[)]`, we get

- t - t 2 2 - t 3 3 - t 4 4 - ⋅⋅⋅ = - ( t + t 2 2 + t 3 3 + t 4 4 + ⋅⋅⋅ ) .

We can now study the power series in parentheses without worrying about keeping track of minus signs. If this series converges for `0<t<1`, so does the logarithmic series for `-1<x<0`. The `n`th tail of the subject series is

Thus, for every positive integer `n` and every fixed number `t` between `0` and `1,` the `n`th tail converges. (In contrast to the error calculation for the exponential series, we did not need to introduce a new variable `r` for the term-to-term ratio in the geometric series; `t` itself is `r`.) Furthermore, the expression

t n + 1 ( n + 1 ) ( 1 - t )

is a bound on the error after summing `n` terms.

Example 4   Determine how close `-0.75-0.75^2/2-0.75^3/3-cdots-0.75^100/100` is to `ln 0.25`.

Solution   The sum here is the first 100 terms of the series for `ln text[(] 1+x text[)]` with `x=-0.75`, so `t=0.75` and `n=100`. The error after summing 100 terms is no bigger than `0.75^101text[/(]101 times 0.25text[)]`, which is approximately `10^-14`. Thus adding 100 terms of the series for `ln 0.25` produces accuracy that exceeds that of most calculators (and the default in most computer algebra systems).

Checkpoint 3