Infinite Sequences and Series

An infinite series 无穷级数 adds infinitely many terms, $\sum_{n=1}^\infty a_n$. Its value is defined as the limit of the partial sums 部分和 $S_N=a_1+a_2+\cdots+a_N$. If $S_N$ approaches a finite number $L$, the series converges 收敛 to $L$; otherwise it diverges 发散. Every convergence question is really a question about the limit of the partial sums.

For the geometric series with $a=1,\ r=\tfrac12$, the partial sums $S_1,S_2,S_3,\dots$ climb toward the limit $\dfrac{a}{1-r}=2$ – that limit is the series' value.

A geometric series 几何级数 $\sum ar^{n}$ has a constant ratio $r$ between terms. It converges exactly when $|r|<1$, and then

This is the one series whose sum you can find exactly, and it underlies power series later in the unit.

Worked example. Sum $3+\tfrac32+\tfrac34+\tfrac38+\cdots$. Here $a=3$ and $r=\tfrac12$ (with $|r|<1$), so the sum is $\dfrac{a}{1-r}=\dfrac{3}{1-\tfrac12}=6$.

A geometric sequence multiplies by the same ratio at each step

If the terms do not shrink to zero, the sum cannot settle: if $\lim_{n\to\infty}a_n\neq0$, the series diverges. This is only a test for divergence – if the terms do go to zero, the test is inconclusive (the series may still diverge, like the harmonic series). Always check this quick test first.

If $a_n=f(n)$ for a positive, decreasing, continuous $f$, then $\sum a_n$ and $\int_1^\infty f(x)\,dx$ both converge or both diverge. The integral test 积分判别法 turns a series question into an improper-integral question, and it is what proves the p-series rule below.

A p-series $\sum \dfrac{1}{n^p}$ converges if $p>1$ and diverges if $p\le1$. The special case $p=1$, $\sum\dfrac1n$, is the harmonic series 调和级数 – it diverges even though its terms go to zero (a famous, must-know fact). The p-series family is the standard yardstick for comparison tests.

Compare an unfamiliar series to a known one (a p-series or geometric series):

• Direct comparison 直接比较: if $0\le a_n\le b_n$ and $\sum b_n$ converges, so does $\sum a_n$; if $a_n\ge b_n\ge0$ and $\sum b_n$ diverges, so does $\sum a_n$.
• Limit comparison 极限比较: if $\lim\dfrac{a_n}{b_n}$ is a finite positive number, the two series do the same thing. This is easier when the terms only behave like a known series.

An alternating series 交错级数 has terms that switch sign, $\sum(-1)^n b_n$. It converges if the $b_n$ are positive, decreasing, and $\lim b_n=0$. This lets series like $\sum\dfrac{(-1)^n}{n}$ converge even though the same terms without the signs (the harmonic series) diverge.

The ratio test 比值判别法 examines $L=\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|$:

• $L<1$: the series converges absolutely;
• $L>1$: it diverges;
• $L=1$: inconclusive.

It is the go-to test for series with factorials or $n$th powers, and it is exactly how you find the radius of convergence of a power series.

Worked example. Test $\displaystyle\sum \frac{n}{2^n}$. The ratio is $\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{n+1}{2^{n+1}}\cdot\dfrac{2^n}{n}=\dfrac{n+1}{2n}\to\dfrac12<1$, so the series converges.

A series converges absolutely 绝对收敛 if $\sum|a_n|$ converges. It converges conditionally 条件收敛 if $\sum a_n$ converges but $\sum|a_n|$ diverges (the classic example is $\sum\dfrac{(-1)^n}{n}$). Absolute convergence is the stronger property; conditional convergence relies on the cancellation of signs.

For a convergent alternating series, the error in stopping at the $N$th partial sum is no larger than the first omitted term:

This simple, powerful bound lets you say how many terms guarantee a desired accuracy.

A Taylor polynomial 泰勒多项式 approximates a function near a center $x=a$ using its derivatives there:

Each added term matches one more derivative, so the polynomial hugs the curve more closely near $a$. Centered at $a=0$ it is a Maclaurin polynomial.

The Maclaurin polynomials of $\sin x$ – $T_1=x$, $T_3$, $T_5$ – each match one more derivative at $0$, so each hugs $\sin x$ over a wider interval before peeling away.

Exam skill: be able to build a Taylor polynomial from a table of derivative values and use it to estimate a function value.

The Lagrange error bound 拉格朗日误差界 bounds how far a Taylor polynomial can be from the true value:

You bound the $(n+1)$th derivative on the interval, then compute – the standard way to prove a Taylor estimate is accurate enough.

A power series 幂级数 $\sum c_n(x-a)^n$ converges for $x$ within a radius of convergence 收敛半径 $R$ of the center $a$. Find $R$ with the ratio test. Then test the two endpoints separately (the ratio test is inconclusive there) to state the full interval of convergence 收敛区间 – including or excluding each endpoint.

Worked example. Find the radius of convergence of $\displaystyle\sum \frac{x^n}{n}$. The ratio test gives $\left|\dfrac{x^{n+1}}{n+1}\cdot\dfrac{n}{x^n}\right|=|x|\dfrac{n}{n+1}\to|x|$, which is $<1$ when $|x|<1$, so $R=1$. Testing the endpoints, $x=-1$ gives the convergent alternating harmonic series and $x=1$ the divergent harmonic series, so the interval is $[-1,1)$.

Extending a Taylor polynomial to infinitely many terms gives a Taylor (or Maclaurin) series. Memorize the key Maclaurin series:

New series come from manipulating these – substituting, differentiating, integrating, or multiplying.

Worked example. Find the Maclaurin series for $e^{x^2}$. Substitute $x^2$ for $x$ in $e^x=\sum\dfrac{x^n}{n!}$:

which converges for all $x$. This substitution trick is far faster than differentiating $e^{x^2}$ six times.

Each extra Maclaurin term hugs the function over a wider range

Because a power series can be differentiated and integrated term by term (within its radius), you can build new series from known ones – e.g. integrate the geometric series for $\dfrac{1}{1-x}$ to get the series for $\ln(1-x)$, or substitute $-x^2$ to get the series for $\dfrac{1}{1+x^2}$. Representing a function as a power series lets you approximate values and integrals that have no elementary antiderivative.

Exam skill: the BC series free-response usually asks you to derive a new Maclaurin series from a known one, find its interval of convergence, and use the alternating-series or Lagrange bound to estimate the error – the capstone skills of the course.

• Test a series for convergence with the right tool: geometric ($|r|<1$, sum $\tfrac{a}{1-r}$), $n$th-term, ratio, integral, comparison, or alternating-series test.
• A geometric infinite sum converges only when $|r|<1$; otherwise it diverges.
• Build a Taylor/Maclaurin series to approximate a function; more terms give a better fit near the centre.
• Know the standard Maclaurin series for $e^x$, $\sin x$, $\cos x$, and $\tfrac{1}{1-x}$.
• Find the radius/interval of convergence with the ratio test, then check the endpoints separately.