Sequences, Geometric and Telescoping Series

Two different limits

A sequence $a_n$ converges to $L$ when its terms approach $L$ as the integer index grows. A series $\sum_{n=1}^\infty a_n$ converges when its partial sums $s_N=\sum_{n=1}^Na_n$ approach a finite limit. The sequence of terms and the sequence of partial sums answer different questions.

A necessary condition for series convergence is $a_n\to0$ , since $a_n=s_n-s_{n-1}$ . It is not sufficient: the harmonic series $\sum1/n$ diverges even though its terms tend to zero. If terms fail to tend to zero, the series diverges immediately.

Visual guide

VISUAL GUIDEDiscrete terms approach a limit

The sequence aₙ = 1/n is defined at integer indices, represented by dots. The terms approach zero; connecting them is unnecessary. A sequence limit concerns terms, whereas a series asks about accumulated sums.

Worked example: geometric accumulation

For $r\ne1$ , multiplying $s_N=1+r+\cdots+r^{N-1}$ by $r$ and subtracting gives $s_N=(1-r^N)/(1-r)$ . If $|r|<1$ , then $r^N\to0$ , so

\sum_{n=0}^\infty ar^n=\frac a{1-r}.

For example, $3+3/2+3/4+\cdots=6$ . If $r=-1/2$ , the sum is $2$ and partial sums alternate around it. At $r=1$ the nonzero terms accumulate without bound; at $r=-1$ partial sums oscillate. Neither endpoint converges for nonzero $a$ .

Worked example: cancellation in partial sums

Because $1/[n(n+1)]=1/n-1/(n+1)$ ,

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

Taking a limit gives sum $1$ , with exact remainder $1/(N+1)$ . The cancellation is justified in a finite sum before taking the limit. Rearranging arbitrary infinite series as though they were finite can change their behavior.

A useful sequence theorem is that every monotone bounded real sequence converges. Nonnegative series have increasing partial sums, so bounding those sums proves convergence. This is the foundation of comparison tests.

Practice

Find $\lim_{n\to\infty}(2n+1)/(n+3)$ .
Sum $\sum_{n=1}^\infty(1/3)^n$ .
Does $\sum_{n=1}^\infty n/(n+1)$ converge?

Show worked solutions

Divide numerator and denominator by $n$ to obtain limit $2$ .
First term $1/3$ , ratio $1/3$ : sum $(1/3)/(2/3)=1/2$ .
No. Its terms tend to $1$ , failing the necessary zero-term test.