P2 4.4 - Infinite geometric sums

Convergence when absolute ratio is below one

Can $8+4+2+1+\cdots$ have a total even though the additions never stop? Its first few partial sums are $8,12,14,15,15.5$: they approach $16$, with a smaller gap each time.

An infinite sum is defined by what happens to these partial sums, not by pretending that there is a final term. A negative ratio can work too. For $8-4+2-1+\cdots$, the partial sums $8,4,6,5,5.5,\ldots$ approach $16/3$ from alternate sides.

[DIAGRAM: asset_name: Partial sums approaching a limit; asset_slug: gp-partial-sum-limits; caption: Both positive and negative ratios can produce convergence; ratio minus one does not.; description: Three panels show partial sums for first term 8 and ratios one half, minus one half and minus one. The first approaches 16 from below, the second alternates towards 16 over 3, and the third alternates between 8 and 0 without a limit.]

Here is why the magnitude matters. From the finite formula,

When $\lvert r\rvert<1$, repeated multiplication makes $r^n$ tend to zero, even if its sign alternates. The partial sums therefore approach a fixed number. Merely having $r<1$ is insufficient: $r=-2$ is less than $1$, but its powers grow in magnitude.

The boundaries are excluded. If $r=1$, then $S_n=na$, which does not tend to a finite limit for $a\ne0$. If $r=-1$, the sums alternate between $a$ and $0$. When $\lvert r\rvert>1$, the nonzero terms grow in magnitude, so the sums do not converge. The all-zero series is the harmless exception: its sum is zero.

For $r=0$, the series is $a+0+0+\cdots$, so the sum is simply $a$.

Check both boundaries, not just whether the ratio is below $1$.

Sum to infinity

Once convergence is established, the limit of the finite-sum formula gives the total directly. The $r^n$ term tends to zero; it is never necessary to find an “infinite last term”.

The opening series has $a=8$ and $r=1/2$, so $S_\infty=8/(1-1/2)=16$. Every finite partial sum is below $16$; the infinite sum names the limiting value.

Change only the sign of the ratio. Convergence survives, but the total decreases because the second term is now subtracted.

Using the formula without checking convergence can produce a meaningless value: $a/(1-r)$ is a finite algebraic expression for many ratios whose infinite series do not converge.

Using finite and infinite sums together

The finite and infinite sums contain the same factor $a/(1-r)$. Use that shared factor to eliminate the first term before solving for the ratio.

For a convergent geometric series,

Consequently, the remaining signed amount after $n$ terms is

This is the sum of the tail beginning with term $n+1$. With negative $r$, it can be negative: a partial sum may lie above the limit, with the remaining terms bringing it back down. These relations follow from the supplied formulas; they are not separately printed in the booklet.

Change only the three-term total from $70$ to $90$. A total above the limiting sum is possible when the terms alternate.

An even power can leave a sign choice. For example, $r^2=1/4$ allows $r=1/2$ and $r=-1/2$; retain both unless other information rules one out. In every case, check the ratio against convergence and the original sums.

Quick Recap

• For a nonzero geometric series, convergence requires $\lvert r\rvert<1$.
• The cases $r=1$ and $r=-1$ do not give finite limiting sums.
• When it converges, $S_\infty=a/(1-r)$ and $S_n=S_\infty(1-r^n)$.
• A negative ratio can make partial sums alternate above and below their limit.

The opening infinite series has sum $16$, although no finite partial sum reaches $16$.