Sequences, Infinite Series & Power Series

A sequence is a function whose input is an integer index; a series is not merely a row of terms but the limit of its finite partial sums. That distinction prevents the most common infinite-process error: manipulating an unproved “sum” as though infinitely many additions had already produced an ordinary number.

Why discrete steps lead to sequences and series #

Many systems change at separate ticks: an algorithm performs iterations, an account updates monthly, and a controller samples a sensor periodically. A sequence records the state or contribution at step $n$. A series answers a different question by accumulating those stepwise contributions.

That distinction is operational: a stopping rule needs the current correction or a bound on the remaining tail, not merely the eventual sum.

Sequence convergence is a tail property #

For

$$a_n=\frac{3n+1}{n+2},$$

subtract the proposed limit $3$:

$$\left\lvert\frac{3n+1}{n+2}-3\right\rvert =\frac{5}{n+2}.$$

Given $\varepsilon>0$, any integer

$$N>\frac5\varepsilon-2$$

ensures the error is below $\varepsilon$. The algebra does more than compute a limit; it produces the index required for a requested accuracy.

Convergent sequences are bounded. The converse is false: $(-1)^n$ is bounded but oscillates. Divergent does not mean unbounded, and monotone does not mean strictly monotone.

Monotonicity plus a bound gives convergence #

A sequence that is eventually nondecreasing and bounded above converges; an eventually nonincreasing sequence bounded below also converges. This monotone convergence theorem is a completeness property of the real numbers: the limit is the supremum or infimum toward which the tail moves.

For recursively defined sequences, the standard proof pattern is:

1. prove the terms stay in an invariant domain;
2. prove a one-sided bound;
3. prove eventual monotonicity;
4. invoke monotone convergence;
5. only then pass to the limit in the recurrence and select the root consistent with the invariant domain.

Solving the fixed-point equation alone is not a convergence proof. A recurrence can have fixed points while its iterates oscillate, diverge, or leave the domain.

Linear recurrences and the corrected Fibonacci formula #

For the Fibonacci convention

$$F_0=0,\qquad F_1=1,\qquad F_{n+2}=F_{n+1}+F_n,$$

try a solution $F_n=r^n$. The characteristic equation is

$$r^2-r-1=0,$$

with roots

$$\varphi=\frac{1+\sqrt5}{2}, \qquad \psi=\frac{1-\sqrt5}{2}.$$

Every solution has form $A\varphi^n+B\psi^n$. The initial conditions give $A=1/\sqrt5$ and $B=-1/\sqrt5$, so

$$F_n=\frac{\varphi^n-\psi^n}{\sqrt5}.$$

Both roots and the factor $1/\sqrt5$ are essential. Check $n=0$ and $n=1$ before trusting a closed form, then substitute it into the recurrence. Since $\lvert\psi\rvert<1$, the second term decays and

$$F_n\sim\frac{\varphi^n}{\sqrt5}.$$

This asymptotic equivalence is not an exact equality, even though rounding $\varphi^n/\sqrt5$ happens to recover $F_n$ for $n\ge0$.

A series is a sequence of partial sums #

The necessary nth-term condition follows immediately:

$$a_N=S_N-S_{N-1}\to S-S=0.$$

Its converse is false. The harmonic terms $1/n$ tend to zero, but grouping shows the partial sums grow without bound:

$$1+\frac12 +\left(\frac13+\frac14\right) +\left(\frac15+\cdots+\frac18\right)+\cdots.$$

Each parenthesized block after the first contains $2^{k-1}$ terms, each at least $1/2^k$, so every block contributes at least $1/2$. Terms becoming small is necessary; they must become small fast enough collectively.

Reference series worth recognizing #

Geometric series #

For a finite sum,

$$S_N=1+r+\cdots+r^N,$$

multiplying by $1-r$ gives

$$(1-r)S_N=1-r^{N+1}.$$

Thus, for $r\ne1$,

$$S_N=\frac{1-r^{N+1}}{1-r}.$$

If $\lvert r\rvert<1$, then $r^{N+1}\to0$ and

$$\sum_{n=0}^{\infty}r^n=\frac1{1-r}.$$

For $\lvert r\rvert\ge1$, the terms do not tend to zero except for no admissible special case producing convergence, so the series diverges.

The tail after degree $N$ is exactly

$$R_N=\frac{r^{N+1}}{1-r},$$

and therefore

$$\lvert R_N\rvert\le\frac{\lvert r\rvert^{N+1}}{1-\lvert r\rvert}.$$

Telescoping series #

Since

$$\frac1{n(n+1)}=\frac1n-\frac1{n+1},$$

the partial sum collapses:

$$\sum_{n=1}^{N}\frac1{n(n+1)} =1-\frac1{N+1}\to1.$$

Always telescope the finite partial sum first. Writing down an infinite cancellation pattern without endpoints can lose the surviving terms or shift an index.

p-series #

$$\sum_{n=1}^{\infty}\frac1{n^p}$$

converges exactly when $p>1$. The boundary $p=1$ is the divergent harmonic series; a tiny positive improvement in exponent changes the conclusion.

Choosing a convergence test #

No single test decides every series. First identify sign structure and dominant scale.

Finite changes to initial terms do not affect convergence, so comparison inequalities and monotonicity conditions only need to hold eventually.

Absolute, conditional, and alternating convergence #

If

$$\sum\lvert a_n\rvert$$

converges, then $\sum a_n$ converges absolutely. Absolute convergence permits regrouping and reordering without changing the sum. A series is conditionally convergent when $\sum a_n$ converges but $\sum\lvert a_n\rvert$ diverges; the alternating harmonic series is the standard example.

For

$$\sum_{n=1}^{\infty}(-1)^{n-1}b_n,$$

if $b_n$ is eventually nonincreasing and $b_n\to0$, then the partial sums alternately over- and undershoot the limit. The remainder after $N$ terms satisfies

$$\lvert S-S_N\rvert\le b_{N+1}.$$

The sign of the first omitted term also gives the direction of the error. But alternation without shrinking-to-zero magnitudes proves nothing, and conditional convergence does not allow arbitrary reordering: a rearrangement can change the sum or destroy convergence.

Error bounds are part of the answer #

A convergence proof says an approximation eventually succeeds; a remainder bound says how many terms are enough.

• Geometric tail: exact formula above.
• Alternating tail: at most the first omitted magnitude under the alternating-series conditions.
• Integral-test tail: for positive decreasing $f$ with $a_n=f(n)$,

  $$\int_{N+1}^{\infty}f(x)\,dx \le S-S_N \le\int_N^{\infty}f(x)\,dx.$$

For $\sum_{n=1}^{\infty}1/n^2$, the integral bound gives

$$0<S-S_N\le\int_N^\infty\frac{dx}{x^2}=\frac1N.$$

Thus $N\ge1000$ guarantees error at most $10^{-3}$. It is conservative but certified.

Power series have a radius—and unresolved endpoints #

A power series centered at $a$ has form

$$\sum_{n=0}^{\infty}c_n(x-a)^n.$$

Exactly one of three cases occurs: it converges only at $a$; it converges for every real $x$; or there is a finite $R>0$ such that it converges absolutely for $\lvert x-a\rvert<R$ and diverges for $\lvert x-a\rvert>R$. At $x=a\pm R$, test each endpoint as an ordinary numerical series.

For

$$\sum_{n=0}^{\infty}\frac{(x-2)^n}{(n+1)3^n},$$

the ratio test yields $\lvert x-2\rvert<3$. At $x=-1$ the result is alternating harmonic and converges; at $x=5$ it is harmonic and diverges. Therefore $R=3$ and the interval is $[-1,5)$.

Within the open interval of convergence, term-by-term differentiation and integration preserve the same radius. Endpoint behavior can change and must be retested.

A reliable workflow #

1. Write the starting index and compute several terms; off-by-one errors often surface immediately.
2. For a sequence, identify a candidate limit and choose direct bounds, squeeze, monotonicity, or a recurrence invariant.
3. For a series, define the finite partial sum before doing algebra.
4. Check $a_n\to0$ first; failure proves divergence.
5. Classify signs and dominant growth, then choose a test whose hypotheses you can prove.
6. Treat ratio/root result $1$ as “no conclusion,” not convergence.
7. Separate absolute from conditional convergence.
8. If using a power series, find the radius and test both endpoints.
9. Attach an error bound whenever the series is used numerically.

Infinite notation becomes trustworthy only after its finite approximants and limiting process are explicit. Partial sums, invariants, and error bounds are the bridge between the symbol $\infty$ and a result that can actually be used.