Infinite Sequences and Series

A sequence $\{a_n\}$ is an ordered list $a_1,a_2,a_3,\dots$ We say $\{a_n\}$ converges to $L$ if $\displaystyle\lim_{n\to\infty}a_n=L$ (a finite number); otherwise it diverges. To find the limit, treat $n$ as a continuous variable and use the same tools as limits at infinity: dominant terms, dividing by the highest power, or L'Hôpital's Rule.

A series is the sum $\displaystyle\sum_{n=1}^{\infty}a_n$. Its value is the limit of the partial sums $S_N=\sum_{n=1}^{N}a_n$: the series converges to $S$ if $\displaystyle\lim_{N\to\infty}S_N=S$. Do not confuse the sequence of terms $a_n$ with the sequence of partial sums $S_N$.

$\displaystyle\sum_{n=0}^{\infty}ar^{n}=a+ar+ar^2+\cdots$ converges if and only if $|r|<1$, and then

If $|r|\ge 1$ it diverges.

If $\displaystyle\lim_{n\to\infty}a_n\neq 0$ (or does not exist), then $\displaystyle\sum a_n$ diverges. Warning: if $\lim a_n=0$, the test is inconclusive --- the series may converge or diverge. It can never prove convergence.

If $f$ is positive, continuous, and decreasing for $x\ge 1$ and $a_n=f(n)$, then

The integral's value is not the sum; it only decides convergence.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ converges if $p>1$ and diverges if $p\le 1$. The case $p=1$, $\sum\frac1n$, is the harmonic series, which diverges.

Suppose $0\le a_n\le b_n$ for all large $n$.

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• If $\sum b_n$ converges, then $\sum a_n$ converges (smaller than convergent).
• If $\sum a_n$ diverges, then $\sum b_n$ diverges (bigger than divergent).

Compare against a known geometric or $p$-series.

If $a_n,b_n>0$ and $\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}=L$ with $0<L<\infty$ (a finite positive number), then $\sum a_n$ and $\sum b_n$ both converge or both diverge. Choose $b_n$ from the dominant-term behavior of $a_n$.

Let $\displaystyle L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$.

Best when $a_n$ contains factorials or $n$th powers.

Let $\displaystyle L=\lim_{n\to\infty}\sqrt[n]{|a_n|}=\lim_{n\to\infty}|a_n|^{1/n}$. Same conclusions as the ratio test ($L<1$ converges, $L>1$ diverges, $L=1$ inconclusive). Best when $a_n$ is raised to the $n$th power, e.g. $\left(\frac{n}{2n+1}\right)^{n}$.

A series $\displaystyle\sum(-1)^{n}b_n$ with $b_n>0$ converges if both hold:

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• $b_n$ is decreasing ($b_{n+1}\le b_n$), and
• $\displaystyle\lim_{n\to\infty}b_n=0$.

If an alternating series meets the AST conditions and converges to $S$, then the error after $N$ terms is bounded by the first omitted term:

$\sum a_n$ converges absolutely if $\sum|a_n|$ converges (this forces $\sum a_n$ to converge). It converges conditionally if $\sum a_n$ converges but $\sum|a_n|$ diverges.

A power series centered at $c$ is $\displaystyle\sum_{n=0}^{\infty}a_n(x-c)^{n}$. There is a radius of convergence $R$ so it converges for $|x-c|<R$ and diverges for $|x-c|>R$. Find $R$ with the ratio test: solve $\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}(x-c)^{n+1}}{a_n(x-c)^{n}}\right|<1$.

The set of $x$ where the series converges is the interval of convergence, $(c-R,\,c+R)$ possibly including one or both endpoints. The ratio test never decides the endpoints ($L=1$ there): substitute each endpoint and test the resulting numeric series separately.

Within the open interval of convergence you may differentiate or integrate term by term, and $R$ stays the same (endpoints may change):

The Taylor series of $f$ centered at $c$ is

A Maclaurin series is the special case $c=0$.

If $P_n(x)$ is the $n$th-degree Taylor polynomial about $c$, the remainder satisfies

where $M$ is any bound on $|f^{(n+1)}|$ between $c$ and $x$. This measures how good the polynomial approximation is.

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• $n$th-term test --- does $a_n\to0$? If not, it diverges. (Always start here.)
• Geometric? form $ar^n$: converges iff $|r|<1$, sum $\frac{a}{1-r}$.
• $p$-series? form $\frac{1}{n^p}$: converges iff $p>1$.
• Factorials or $n$th powers? Use the ratio (or root) test.
• Looks like a $p$-series/geometric? Use direct or limit comparison.
• Positive, decreasing, easily integrated? Use the integral test.
• Alternating $(-1)^n b_n$? Use the AST; then check $\sum|a_n|$ for absolute vs. conditional.

Substitute into a known Maclaurin series instead of computing derivatives. Replace $x$ by any expression:

After finding $R$ by the ratio test, each endpoint is tested separately and independently. A single power series can produce any of $(c-R,c+R)$, $[c-R,c+R)$, $(c-R,c+R]$, or $[c-R,c+R]$. Typical tools at an endpoint: the $p$-series test, the $n$th-term test, or the alternating series test. Never assume symmetry.