7. Sequences, finite sums and convergent series

Learning goals and definitions

A sequence $(a_n)_{n\ge1}$ is an ordered list of numbers. It converges to $L$ if for every $\varepsilon>0$ there is a positive integer $N\in\mathbb Z_{\ge1}$ such that $n\ge N$ implies $|a_n-L|<\varepsilon$. The series $\sum_{n=1}^{\infty}a_n$ is defined through its partial sums $S_N=\sum_{n=1}^N a_n$: it converges precisely when $S_N$ has a finite limit. Terms and partial sums are different sequences.

An arithmetic progression has $a_n=a_1+(n-1)d$, and pairing the first and last terms gives $S_N=N(a_1+a_N)/2$. A geometric progression has $a_n=ar^{n-1}$, with exact finite sum

$$S_N=\frac{a(1-r^N)}{1-r}\quad(r\ne1),\qquad S_N=Na\quad(r=1).$$

Derivation. Subtract $rS_N=ar+\cdots+ar^N$ from $S_N=a+\cdots+ar^{N-1}$. All interior terms cancel, leaving $(1-r)S_N=a-ar^N$. This is a finite algebraic identity requiring no convergence assumption.

Theorem: the geometric series and its tail

If $|r|<1$, then $\sum_{n=1}^{\infty}ar^{n-1}=a/(1-r)$ and the error after $N$ terms is $ar^N/(1-r)$. For $a\ne0$ and $|r|\ge1$, the series does not converge.

Proof. For $|r|<1$, $r^N\to0$, so the finite formula has the claimed limit and exact error. For $|r|\ge1$, the terms do not tend to zero. Any convergent series must have $a_n=S_n-S_{n-1}\to0$, so convergence is impossible. The special zero series $a=0$ converges for every $r$.

Terms tending to zero are necessary but not sufficient. The harmonic series $\sum1/n$ diverges: the block $n=2^{k-1}+1,\ldots,2^k$ has $2^{k-1}$ terms each at least $1/2^k$, so every block contributes at least $1/2$. Its partial sums are therefore unbounded.

A telescoping sum should also be evaluated finitely before taking limits. For example

$$\sum_{n=1}^N\frac1{n(n+1)}=\sum_{n=1}^N\left(\frac1n-\frac1{n+1}\right)=1-\frac1{N+1}.$$

Thus the infinite sum is $1$. Infinite rearrangements are not justified by the finite cancellation alone; the displayed order is part of this calculation.

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.

Worked examples

An error target. Approximate $\sum_{n=0}^{\infty}3(1/4)^n=4$ by terms $n=0,\ldots,N-1$. The omitted tail is $4(1/4)^N$. To make it less than $10^{-3}$, require $4^N>4000$. Since $4^5=1024$ and $4^6=4096$, six terms suffice and five do not.

A repeating decimal. The decimal $0.272727\ldots$ means $\sum_{k=1}^{\infty}27\cdot100^{-k}$. Its sum is $(27/100)/(1-1/100)=27/99=3/11$. The geometric definition makes the infinite notation precise.

Graded practice

1. Foundation. Find $\sum_{n=1}^{20}(3n-1)$.
2. Core. Determine the sum of $\sum_{n=1}^{\infty}2(-1/3)^{n-1}$ and bound the absolute error after four terms.
3. Core. Evaluate $\sum_{n=1}^{\infty}1/((n+1)(n+2))$.
4. Proof. Explain why $\sum n/(n+1)$ diverges, then give a divergent series whose terms nevertheless tend to zero.

Full solutions

1. Use $\sum_{n=1}^{20}n=20\cdot21/2=210$. The sum is $3\cdot210-20=610$.
2. The ratio has magnitude less than one, so the sum is $2/(1+1/3)=3/2$. The exact tail is $[2(-1/3)^4]/(1+1/3)=1/54$, whose absolute value is also $1/54$.
3. Write $1/((n+1)(n+2))=1/(n+1)-1/(n+2)$. The first $N$ terms sum to $1/2-1/(N+2)$, hence the limit is $1/2$.
4. The terms $n/(n+1)$ tend to $1$, violating the necessary zero-term condition. The harmonic series has terms $1/n\to0$ but diverges by the block estimate above. Thus the condition cannot be reversed.

Summary and term definitions

A partial sum is a finite accumulated sum. A series converges when its partial sums converge, while a tail is the omitted remainder after truncation. A telescoping series simplifies by cancellation of adjacent terms in finite partial sums. Identify the exact finite identity first, then justify the limit; never substitute an infinite number of terms into a formula without checking its hypotheses.