Testing a Point with a Geometric Bound
A power series can be viewed at a particular value of its variable as a sequence of terms. At that point, the size of each term depends on both the coefficient and the distance from the center. Geometric bounds let us estimate those terms without needing to find a sum for the series.
The direction of the bound matters. An upper bound can show that the terms are no larger than the terms of a convergent geometric series. A lower bound can show that the terms fail to approach zero, which rules out convergence. Neither conclusion follows merely from having some geometric-looking expression: the inequality and its ratio must be checked at the specified point.
Write \(t=|x-c|\) for the distance from the point to the center. The point is inside the radius of convergence \(R\) when \(t<R\), and outside it when \(t>R\). The radius results established earlier say that a power series converges absolutely inside its radius and diverges outside it. Geometric bounds provide a direct way to verify what happens at a specified point.
Upper Bounds Certify Convergence
Suppose that, for all sufficiently large \(n\), the coefficients satisfy \(|a_n|\le Mq^n\), where \(M>0\) and \(q>0\). At a point whose distance from the center is \(t\), this gives
If \(qt<1\), the right-hand side decreases geometrically. The tail of the power series is then bounded by a convergent geometric series. The finitely many terms before the coefficient bound begins do not affect convergence.
Why the Upper-Bound Test Works
Here is the complete comparison argument. Put \(r=qt\), and suppose \(r<1\). For every \(n\ge N\), the magnitude of the \(n\)th term is at most \(Mr^n\). For any integer \(K\ge N\), the sum of the magnitudes from index \(N\) through \(K\) is therefore at most
More importantly, the tail beginning at an index \(J\ge N\) is bounded by \(Mr^J/(1-r)\). Since \(0\le r<1\), this bound tends to zero as \(J\) increases. Thus the partial sums of the absolute values satisfy the Cauchy criterion and converge. Consequently, the original power series converges absolutely at \(x\). This argument explains why an upper geometric bound with ratio strictly less than one is enough.
Worked Example: Bounding Coefficients with Linear Growth
Worked Example: Bounding Coefficients with Linear Growth
Test the power series \(\sum_{n=0}^{\infty}\frac{n+1}{5^n}(x-1)^n\) at \(x=2\) using a geometric coefficient bound.
The center is \(c=1\), so the distance to the test point is
For every integer \(n\ge0\), \(n+1\le2^n\). To verify this, equality holds at \(n=0\) and \(n=1\). If \(n+1\le2^n\), then \(n+2\le2(n+1)\le2^{n+1}\), proving the inequality for all \(n\) by induction. Thus \(|a_n|=(n+1)/5^n\le(2/5)^n\). At \(x=2\),
The geometric ratio is \(2/5<1\). The sum of the comparison series is \(\sum_{n=0}^{\infty}(2/5)^n=1/(1-2/5)=5/3\). Therefore the power series converges absolutely at \(x=2\), and its sum of absolute values is at most \(5/3\).
Check: The coefficient inequality holds for every index, the point distance is \(1\), and multiplying the coefficient bound by \(1^n\) leaves the ratio \(2/5\), strictly less than \(1\).
Lower Bounds Can Prove Divergence
An upper bound cannot show divergence: a large upper estimate might simply be too weak. For a divergence argument, a lower bound is useful. If \(|a_n|\ge mq^n\) eventually, where \(m>0\), then at distance \(t\),
If \(qt>1\), this lower bound grows rather than tending to zero. Even when \(qt=1\), it stays at least \(m\). In either case the terms do not approach zero, so the series cannot converge. This uses the necessary term condition for convergence: the terms of a convergent series must tend to zero.
Worked Example: Comparing Points Inside and Outside
Worked Example: Comparing Points Inside and Outside
Consider \(\sum_{n=0}^{\infty}3^n(x-2)^n\). Test the points \(x=2.2\) and \(x=2.5\).
Here \(a_n=3^n\), so both an upper and a lower geometric bound hold exactly: \(|a_n|=3^n\). The center is \(c=2\).
At \(x=2.2\), the distance is \(t=|2.2-2|=0.2\), and the term magnitudes are
The ratio \(0.6\) is less than \(1\), so this is a convergent geometric series. Its sum is \(1/(1-0.6)=2.5\). The power series therefore converges absolutely at \(x=2.2\).
At \(x=2.5\), the distance is \(t=|2.5-2|=0.5\). The term magnitudes are
These terms do not approach zero; for example, they grow because their ratio is \(1.5>1\). Thus the power series diverges at \(x=2.5\). Since its coefficient growth is exactly \(3^n\), its radius is \(R=1/3\): the first point is at distance \(0.2<1/3\), and the second is at distance \(0.5>1/3\).
Check: The two distances and products are \(3(0.2)=0.6<1\) and \(3(0.5)=1.5>1\), matching the convergence and divergence conclusions.
When a Geometric Bound Is Inconclusive
The strict inequality in the upper-bound test is essential. If \(qt=1\), the estimate only says that the term magnitudes are at most \(M\). A sequence bounded above by a constant need not have a summable series, so this bound does not establish convergence. Likewise, a lower bound with \(qt<1\) does not establish divergence: a lower estimate that tends to zero does not prevent the actual terms from tending to zero.
Bounds are also estimates, not necessarily exact descriptions of the coefficients. A poor choice of \(q\) may make \(qt\ge1\) even though the series does converge at the point. That does not contradict the convergence theorem; it means only that this particular upper bound cannot certify it. A sharper bound, or a different method, may be needed.
Worked Example: A Polynomial Factor Still Allows a Geometric Bound
Test \(\sum_{n=0}^{\infty}\frac{n^2+1}{10^n}x^n\) at \(x=2\).
For all \(n\ge0\), \(n^2+1\le4^n\). At \(n=0\) and \(n=1\), this reads \(1\le1\) and \(2\le4\). For \(n\ge1\), if the inequality holds at \(n\), then
because \(4(n^2+1)-((n+1)^2+1)=3n^2-2n+2>0\). Induction proves the bound. Therefore \(|a_n|\le(4/10)^n\). The point is at distance \(t=|2-0|=2\), so
The comparison ratio is \(4/5<1\); the geometric comparison series sums to \(1/(1-4/5)=5\). Hence the power series converges absolutely at \(x=2\), with sum of absolute values at most \(5\).
Check: The coefficient estimate is valid for every index, and the point distance changes its ratio from \(4/10\) to \((4/10)(2)=4/5\), which remains below \(1\).
A Common Mistake: Treating an Upper Bound as a Lower Bound
Suppose a calculation gives \(|a_n(x-c)^n|\le 2^n\). A common incorrect conclusion is: “The terms grow like \(2^n\), so the series diverges.” The inequality does not say that. It says the terms are no larger than \(2^n\); they might be much smaller. For instance, \(0\le(1/4)^n\le2^n\) for every \(n\ge0\), even though \((1/4)^n\) tends to zero.
The correction is to use the direction of the comparison properly. An upper bound proves convergence only when the comparison terms form a convergent geometric series. To prove divergence from a bound, one needs a suitable lower bound that prevents the terms from tending to zero.
Check Your Understanding
For each question, focus on the coefficient inequality and its geometric ratio at the specified point.
- If \(|a_n|\le5(2/3)^n\), for which distances \(t=|x-c|\) does this bound prove absolute convergence?
- If \(|a_n|\ge(4/3)^n\) eventually, what does the lower-bound test conclude at a point with \(t=3/4\)?
- Why does the estimate \(|a_n(x-c)^n|\le7\) not by itself prove convergence?
- For coefficients \(a_n=(n+1)/8^n\), use \(n+1\le2^n\) to find distances \(t\) for which the upper-bound test certifies convergence.
- If an upper-bound test produces geometric ratio exactly \(1\), what can you conclude from that bound alone?