Defining Convergent and Divergent Infinite Series

Can infinitely many moves cover a finite distance?

• A robot moves $1$ metre, then $1/2$ metre, then $1/4$ metre, and keeps halving its next move. The total distance can approach $2$ metres even though there is always another move.
• An infinite series 无穷级数 adds the terms of a sequence: $\sum_{n=1}^{\infty}a_n=a_1+a_2+\cdots$. To give this a value, study its finite running totals.

Build the partial sums

• The $N$th partial sum 部分和 is $S_N=\sum_{n=1}^{N}a_n=a_1+\cdots+a_N$. For the robot, $S_1=1$, $S_2=3/2$, $S_3=7/4$ and $S_4=15/8$.
• The terms $a_1,a_2,\ldots$ and the partial sums $S_1,S_2,\ldots$ are different sequences. Here $a_3=1/4$, but $S_3=7/4$.

Define convergence using a limit

• A series converges 收敛 to $L$ exactly when $\lim_{N\to\infty}S_N=L$ exists and is finite. Then write $\sum_{n=1}^{\infty}a_n=L$.
• Otherwise the series diverges 发散. A few totals or a finite graph can suggest a pattern, but do not prove an infinite limit.

Prove the robot's total

• Its first $N$ moves add to $S_N=1+1/2+\cdots+1/2^{N-1}=2(1-2^{-N})$. Subtracting half this finite sum from itself gives $S_N/2=1-2^{-N}$.
• Hence $2-S_N=2^{1-N}\to0$, so $S_N\to2$. Each finite total is below $2$; the infinite sum is $2$ even though no finite total equals it.

The exact formula explains the pattern shown by these first nine totals.

Oscillation can still converge

• For $1-1/2+1/4-1/8+\cdots$, the partial sums are $1,1/2,3/4,5/8,\ldots$. They move above and below a common limit.
• Their formula is $S_N=\dfrac23\left(1-(-1/2)^N\right)$. Since $(-1/2)^N\to0$, the series converges to $2/3$. Convergent partial sums need not increase or decrease throughout.

Bounded totals can fail to converge

• For $1-1+1-1+\cdots$, odd partial sums equal $1$ and even partial sums equal $0$. These two groups approach different values, so there is no single limit.
• Every total lies between $0$ and $1$, yet the series diverges. Staying bounded or alternating signs alone does not prove convergence.

Divergence need not mean oscillation

• For $1+1+1+\cdots$, $S_N=N\to+\infty$. There is no finite series sum, so this series diverges too.
• Compare the reasons: unbounded growth for this series, and two distinct odd/even limits for $1-1+1-1+\cdots$. Both fail the same finite-limit definition.

Recover terms from running totals

• Set $S_0=0$. Then $a_1=S_1$ and $a_N=S_N-S_{N-1}$ for $N\ge2$: subtract the previous total to find the new contribution.
• Suppose $S_N=4-\dfrac1{N+1}$ for $N\ge1$. The series sums to $4$, while $a_1=7/2$ and $a_N=\dfrac1N-\dfrac1{N+1}=\dfrac1{N(N+1)}$ for $N\ge2$.

Terms must tend to zero

• If $S_N\to L$, then $S_{N-1}\to L$, so $a_N=S_N-S_{N-1}\to L-L=0$. This is a necessary condition for convergence.
• This says the terms tend to zero, not that their magnitudes must decrease at every step. A nonzero term limit rules out convergence; a zero term limit alone does not prove it.

A zero term limit is not enough

• The harmonic series $\sum_{n=1}^{\infty}1/n$ has terms tending to zero but divergent partial sums. In each block $n=2^{k-1}+1,\ldots,2^k$, the terms add to at least $1/2$.
• For example, $1/3+1/4\ge1/2$ and $1/5+\cdots+1/8\ge1/2$. Infinitely many such blocks force unbounded totals. Compare the robot's series: its terms also tend to zero, but its totals tend to $2$.

Try fresh partial-sum models

• For a series with $S_N=2-\dfrac3{N+1}$ for $N\ge1$, find $a_1$, $a_2$, a formula for $a_N$ when $N\ge2$, and the series sum. State why you use a separate first term.
• Decide what happens when $S_N=3+\dfrac{(-1)^N}{N}$, and when $S_N=3+(-1)^N$. Explain using limits of the totals, rather than just the word “oscillates”.

Choose the sequence you need

• To find the value of a series, take the limit of $S_N$, not $a_N$. If only $a_N$ is given, build the totals or use a suitable convergence argument.
• A finite partial-sum limit gives convergence. Unbounded totals or different odd/even limits give divergence. Terms tending to zero and bounded totals are each insufficient on their own.