Series depth: finite identities, models and certified tails

Finite identities, models and tails

A memorized infinite-sum formula does not answer three earlier questions: which finite algebraic identity is being used, whether its endpoint limit exists, and how much remains after a finite computation. We develop those layers separately, then apply them to a recurrence and a waiting-time distribution. Prerequisites are finite algebra, the geometric-sum formula, elementary logarithms and the definition of series convergence; no differentiation of infinite series is used.

Proof bridge: finite sums, geometric limits and comparison

In the geometric identities below, $a,r$ may be real or complex; put $\rho=|r|$. Signed remainders are understood as complex remainders when appropriate. For complex sums, convergence means convergence in modulus.

We take the Archimedean property and completeness of the real numbers as basic properties of $\mathbb R$. The Archimedean property says that for every real $M$ there is a positive integer $N\in\mathbb Z_{\ge1}$ with $N>M$. Completeness says that every nonempty real set bounded above has a least upper bound. Induction is used only for finite integer arguments. No infinite cancellation or rearrangement is assumed.

Geometric powers tend to zero. Let $0\le\rho<1$. If $\rho=0$, then $\rho^n=0$ for $n\ge1$. Otherwise put $t=1/\rho-1>0$. The inequality $(1+t)^n\ge1+nt$ follows by induction: multiplication of $1+nt$ by $1+t$ gives $1+(n+1)t+nt^2\ge1+(n+1)t$. Therefore $0\le\rho^n\le1/(1+nt)$. For any $\varepsilon>0$, the Archimedean property supplies $N$ with $1+Nt>1/\varepsilon$; then $n\ge N$ gives $\rho^n<\varepsilon$. Thus $r^n\to0$ whenever $|r|<1$.

For $|r|<1$ and the finite geometric sum $S_N=a(1-r^N)/(1-r)$, direct subtraction gives

$$\left|S_N-\frac a{1-r}\right|=\frac{|a|\,|r|^N}{|1-r|}\longrightarrow0.$$

This proves the infinite sum and its exact signed remainder $a r^N/(1-r)$ directly from the definition of convergence, including negative $r$. If $a=0$, every partial sum is zero for every real or complex $r$. If $a\ne0$ and $|r|\ge1$, the terms have magnitude at least $|a|$ and do not tend to zero.

Why convergent series have terms tending to zero. If $S_n\to S$, then

$$|a_n|=|S_n-S_{n-1}|\le|S_n-S|+|S_{n-1}-S|.$$

For any $\varepsilon>0$, both terms on the right are less than $\varepsilon/2$ for all sufficiently large $n$. Hence $a_n\to0$. Also, every convergent real sequence is bounded: eventually its distance from its limit is less than one, and its finitely many earlier values are bounded. These facts justify the geometric-series divergence assertion above.

Harmonic divergence by finite blocks. For each integer $k\ge1$, the finite block $n=2^{k-1}+1,\ldots,2^k$ has $2^{k-1}$ terms, each at least $1/2^k$. Consequently

$$\sum_{n=1}^{2^K}\frac1n =1+\sum_{k=1}^K\sum_{n=2^{k-1}+1}^{2^k}\frac1n \ge1+\frac K2.$$

These partial sums are unbounded, so the harmonic series cannot converge, although $1/n\to0$ by the Archimedean property. All grouping above is finite; it does not appeal to a theorem about rearranging infinite series.

Comparison and absolute convergence. First, a nondecreasing real sequence $(s_n)$ bounded above converges to $L=\sup\{s_n:n\ge1\}$. Indeed, for each $\varepsilon>0$ some $s_N>L-\varepsilon$, or else $L-\varepsilon$ would be an upper bound; for $n\ge N$, $L-\varepsilon<s_n\le L$.

If $0\le u_n\le b_n$ from some index onward and $\sum b_n$ converges with $b_n\ge0$, the tail partial sums of the $u_n$ are nondecreasing and bounded by a fixed finite initial contribution plus a bound for the $b_n$ partial sums. The preceding supremum argument proves that $\sum u_n$ converges. For real $a_n$, if $\sum|a_n|$ converges, apply this argument to $a_n^+=(|a_n|+a_n)/2$ and $a_n^-=(|a_n|-a_n)/2$, each between zero and $|a_n|$. Their partial sums converge, so their difference $\sum_{n=1}^N a_n$ converges: the error in the difference is at most the sum of the two individual errors. For complex $a_n$, apply the real result to real and imaginary parts, whose absolute values are at most $|a_n|$. If these component sums have limits $U,V$, the complex partial sums tend to $U+iV$ because $|z-(U+iV)|\le|\operatorname{Re}z-U|+|\operatorname{Im}z-V|$. This proves absolute convergence implies convergence, and justifies comparison with a convergent geometric majorant. A finite number of initial terms does not affect convergence because it only adds a fixed finite constant to later partial sums.

1. Index shifts are finite algebra

For a positive integer $N$, define $S_N(r)=\sum_{k=1}^Nkr^k$. Writing $rS_N$ changes both the exponent and its associated coefficient:

$$S_N=r+2r^2+\cdots+Nr^N, \qquad rS_N=r^2+2r^3+\cdots+Nr^{N+1}.$$

Subtracting gives

$$(1-r)S_N=\sum_{k=1}^Nr^k-Nr^{N+1}.$$

The last term is a boundary term. It must not disappear merely because it is outside the original index range. If $r\ne1$, insert the finite geometric identity and obtain

$$S_N(r)=\frac{r-(N+1)r^{N+1}+Nr^{N+2}}{(1-r)^2}.$$

If $r=1$, division by $1-r$ is unavailable, and the original sum instead gives $N(N+1)/2$. The finite identity is valid for every other real or complex $r$, including those for which the infinite series diverges. For example it is still an exact formula at $r=2$.

Why this is a general method. Multiplying a weighted sum by its ratio shifts the same weights by one place. Subtraction replaces a linear weight $k$ by the simpler difference $k-(k-1)=1$. More complicated polynomial weights can be reduced by further finite differences. Each subtraction also creates endpoint terms, so a derivation should write those terms explicitly before collecting powers.

2. Passing to infinity requires another argument

For $|r|<1$, $Nr^N\to0$. To justify the extra factor $N$, let $\rho=|r|$ with $0<\rho<1$ and choose $q$ with $\rho<q<1$. For all sufficiently large $n$, $\rho(1+1/n)\le q$. Therefore the positive terms $a_n=n\rho^n$ satisfy $a_{n+1}\le qa_n$ from that point onward. Repeated comparison bounds them by a geometric sequence tending to zero. The case $r=0$ is immediate.

Taking limits of the proved finite identity now yields

$$\sum_{k=1}^{\infty}kr^k=\frac r{(1-r)^2},\qquad |r|<1.$$

For $|r|\ge1$, the individual terms $kr^k$ do not approach zero, so there is no convergent infinite series. A finite formula never grants permission to insert an infinite index outside its convergence range.

For a signed or complex ratio with $|r|<1$, the exact tail is

$$\sum_{k=N+1}^{\infty}kr^k =\frac{r^{N+1}[(N+1)-Nr]}{(1-r)^2}.$$

One derivation writes $k=N+j$ and separates $\sum_{j\ge1}(N+j)r^{N+j}$ into a geometric sum and the weighted sum just established. A certified absolute upper bound uses $\rho=|r|$:

$$\left|\sum_{k=N+1}^{\infty}kr^k\right| \le\frac{\rho^{N+1}[(N+1)-N\rho]}{(1-\rho)^2}.$$

This estimate may be larger than the signed error because it does not rely on cancellation.

3. A recurrence is a model with a clock

Suppose a reservoir contains $B_n$ units immediately after addition number $n$. Between additions, a fraction $q$ remains, initially with $0<q<1$, and then $d\ge0$ new units are added. Thus

$$B_{n+1}=qB_n+d.$$

Expanding a few steps reveals a pattern, but induction proves it:

$$B_n=q^nB_0+d\sum_{j=0}^{n-1}q^j =q^nB_0+\frac{d(1-q^n)}{1-q}.$$

The base $n=0$ holds with an empty sum. Multiplication by $q$ followed by adding $d$ shifts the finite sum to the next case. Therefore $B_n$ tends to $B_*=d/(1-q)$, and subtracting the fixed point gives the exact error $B_n-B_*=q^n(B_0-B_*)$.

Worked model and interpretation. With $q=3/4,d=2,B_0=0$, $B_n=8(1-(3/4)^n)$, so eight is a limiting level, not a level attained after finitely many additions. To be within $0.1$ of it requires $8(3/4)^n<0.1$, or $n>\ln80/\ln(4/3)$. The least integer is sixteen: the exact comparisons $80\cdot3^{15}>4^{15}$ and $80\cdot3^{16}<4^{16}$ certify the threshold without rounding logarithms.

If addition happens before retention, the recurrence becomes $C_{n+1}=q(C_n+d)$, with limiting level $qd/(1-q)$. The two models have different meanings even though they use the same physical constants. Decide when the quantity is observed before writing the recurrence. The degenerate case $q=0$ is checked directly: after-retention addition gives $B_n=d$ for $n\ge1$, while before-retention addition gives $C_n=0$ for $n\ge1$. This avoids using an ambiguous numeric $0^0$ in a formula at $n=0$.

4. A probability synthesis: waiting until first success

Take independent trials with common success probability $p$, first assuming $0<p<1$. Let $X$ be the number of trials up to and including the first success. For $k\ge1$,

$$P(X=k)=(1-p)^{k-1}p.$$

Here independence means that every finite collection of trial outcomes has its joint probability equal to the product of its marginals. Except on the null event specified below, $X=k$ means $k-1$ failures followed by one success, which gives the displayed probability. For every positive integer $m$, the event of never succeeding is contained in the event of failing the first $m$ trials, whose probability is $(1-p)^m$. Since this bound tends to zero, never succeeding has probability zero. We define $X=1$ on that null event, without changing its distribution. The geometric identity gives $\sum_{k=1}^m p(1-p)^{k-1}=1-(1-p)^m\to1$, so the countable distribution is normalized. For a nonnegative integer-valued random variable, its expectation is defined by the nonnegative series $\sum_k kP(X=k)$, possibly infinite. In this case the proved weighted-sum formula shows it is finite:

$$E[X]=\sum_{k\ge1}kp(1-p)^{k-1}=\frac1p.$$

At $p=1$, $X=1$ with certainty and the same answer holds directly. At $p=0$, a first success never occurs; it is invalid to treat $1/p$ as a finite expectation or claim that the displayed weights still sum to one.

5. Error diagnosis: term size is not tail size

A stopping rule “the next term is below the tolerance” is not generally valid. For a positive geometric tail beginning with term $a$ and ratio $r\in(0,1)$, the whole tail is $a/(1-r)$. If $r=99/100$ and the first omitted term is $10^{-4}$, the omitted total is $10^{-2}$, one hundred times that term. A rigorous stopping rule controls the sum of all omitted terms, using a derived tail bound.

Alternating signs alone are also insufficient. Let $b_n=(2+(-1)^n)/n>0$. Although $b_n\to0$, the series $\sum_{n\ge1}(-1)^nb_n$ diverges: its $k$th consecutive pair is

$$\frac3{2k}-\frac1{2k-1}\ge\frac1{2k},$$

so its even partial sums dominate a divergent harmonic multiple. The usual alternating-series test needs eventual monotone decrease of the positive magnitudes as well as their convergence to zero. Here the magnitudes repeatedly jump upward; ignoring that hypothesis is the first invalid step.

Graded synthesis and diagnostics

1. Derivation. Starting from the finite weighted identity, derive a formula for $\sum_{k=0}^{N-1}(k+1)r^k$ when $r\ne1$, and handle $r=1$ separately. Explain the change of indices.
2. Model comparison. In the reservoir model, retain fraction $q$ after adding $d$. Derive the exact formula for $C_n$ from arbitrary $C_0$ and compare its limit with the add-after-retention model.
3. Probability. For the first-success model with $0<p<1$ and a positive integer $N$, compute $P(X>N)$ directly and explain why the tail probability and the expectation are different quantities.
4. Certified computation. Let $\varepsilon>0$. A positive geometric series has ratio $4/5$. What condition on its first omitted term is sufficient to make the entire omitted sum less than $\varepsilon$? Must that term itself be less than $\varepsilon$?
5. Proof audit. A student removes the last term in $(1-r)S_N=\sum_{k=1}^Nr^k-Nr^{N+1}$ before taking limits, saying “an endpoint has no effect.” Identify when the final infinite formula may still be correct and why the argument is invalid as written.
6. Challenge. Prove the pair lower bound for $\sum(-1)^nb_n$ above, then explain why a subsequence of partial sums tending to infinity is sufficient to rule out convergence of the entire partial-sum sequence.

Full solutions

1. Let $j=k+1$. The sum is $\sum_{j=1}^N jr^{j-1}$. For $r\ne0,1$ it is $S_N(r)/r$, hence $[1-(N+1)r^N+Nr^{N+1}]/(1-r)^2$. At $r=0$, the original sum equals one, and the same displayed expression gives one for $N\ge1$, so it remains valid. At $r=1$, sum the integers from one through $N$, giving $N(N+1)/2$. This handles the point at which dividing by $r$ was not legitimate.
2. The recurrence is $C_{n+1}=qC_n+qd$. For $0<q<1$ and $n\ge0$, induction gives $C_n=q^nC_0+qd(1-q^n)/(1-q)$, tending to $qd/(1-q)$. For $q=0$, the initial value is $C_0$ and $C_n=0$ for $n\ge1$, so the limit is zero. The add-after-retention limit is $d/(1-q)$, larger by $d$ when $0\le q<1$. At the observation instant, the new addition is retained only in the first model's next period but immediately in the second model's current period.
3. The event $X>N$ means the first $N$ trials all fail, so its probability is $(1-p)^N$ by independence. It is a dimensionless chance of exceeding a threshold, whereas $E[X]=1/p$ is a probability-weighted average number of trials. One cannot replace one by the other or infer a deterministic waiting-time guarantee from the expectation.
4. The whole tail is five times its first term $a$, so $a<\varepsilon/5$ is sufficient and, for this positive exact geometric tail, necessary. Certainly $a<\varepsilon$ is necessary for a total below $\varepsilon$, but it is not sufficient: a term between $\varepsilon/5$ and $\varepsilon$ passes that weaker test while its tail does not.
5. The endpoint term is part of the exact finite identity and usually nonzero. It may be removed only after proving $Nr^{N+1}\to0$, which holds for $|r|<1$. In that case the final infinite formula is correct, but the missing estimate is a logical gap. For $r=2$, the endpoint term grows and cannot be discarded; the purported infinite sum would contradict the zero-term necessary condition.
6. Subtract $1/(2k)$ from the pair to obtain $1/k-1/(2k-1)=(k-1)/[k(2k-1)]\ge0$. Thus even partial sums are at least $\sum_{k=1}^N1/(2k)$, unbounded. Every subsequence of a convergent real sequence must converge to the same finite limit; this divergent subsequence therefore rules out convergence of the full sequence.

Summary and terms

An index shift changes labels while preserving the specified finite terms. A boundary term is left by shifting or cancellation at an endpoint. A recurrence describes a quantity's update in a chosen observation clock. A certified tail bounds every omitted term together, not merely the next one. Finite identities, limiting arguments, model assumptions and computational error bounds are distinct layers and should be justified in that order.