Infinite Series

An infinite series, or series, is the sum of an sequence $$a_1 + a_2 + a_3 + \cdots + a_n + \cdots$$

Example: $$1+{1\over 2}+{1\over 4}+{1\over 8}+{1\over 16}+\cdots$$

For series, we need to invest the sum of the infinity many numbers.

Consider its $n$th partial sum $$s_n=a_1 + a_2 + a_3 + \cdots + a_n$$ which is a sequence.

Given a series $$a_1+a_2+a_3+\cdots+a_n+\cdots$$ The number $a_n$ is the $n$th term of the series. If the sequence of partial sums converges to a limit $L$, we say that the series converges and its sum is $L$. In this case, we aslo write $$a_1+a_2+a_3+\cdots+a_n+\cdots=\sum_{n=1}^\infty a_n=L$$ If the sequence of partial sums of the series $s_n$ does not converge, we say that the series diverges.

There are different notations for a serie: $$\sum_{n=1}^\infty a_n,\quad \sum_{k=1}^\infty a_k,\quad \sum a_n$$

Geometric Series ($a\neq 0$ and $r$ are fixed)

• $r=1$

$$\sum_{n=1}^{N}ar^{n-1}=Na$$

• $r\neq1$

Let $s_N=\sum\limits_{n=1}^{N}ar^{n-1}$, then we have $$r\cdot s_N=\sum_{n=1}^{N}ar^{n}=s_N-a+ar^N.$$ Thus we have $$s_N =\sum_{n=1}^{N}ar^{n-1}= {a-ar^N\over 1-r}$$

Examples:

• $$\sum_{n=0}^\infty {(-1)^n5\over 4^n}$$

We have $a=5$ and $r=-{1\over 4}$, then the series converges to ${a\over 1-r}=4$.

Telescoping Example: $\sum\limits_{n=1}^\infty {1\over n(n+1)}$

The $n$th-Term Test for a Divergent Series

If $\sum_{n=1}^\infty a_n$ converges, then $a_n\rightarrow 0$. (This is a necessary condition for convergence.)

$\sum\limits_{n=1}^\infty a_n$ diverges if $\lim_{n\rightarrow \infty}a_n$ fails to exists or is different from zero. (This is a sufficient condition for divergence.)

Examples: Divergence or Convergence

• $\sum\limits_{n=1}^\infty n^2$ divergence because $a_n\rightarrow \infty$.

• $\sum\limits_{n=1}^\infty (-1)^{n+1}$ divergence because $a_n$ does not convergence.

• $\sum\limits_{n=1}^\infty {-n\over 2n+5}$ divergence because $a_n$ converges to $-{1\over 2}$.

$\lim\limits_{n\rightarrow \infty}a_n=0$ does not mean $\sum\limits_{n=1}^\infty a_n$ converges

Example: $$\sum_{n=1}^\infty {1\over n}$$

Combing Series

If $\sum a_n=A$ and $\sum b_n=B$ are convergent series, then

• Sum rule: $\sum(a_n+b_n)=\sum a_n+\sum b_n = A+B$

• Difference rule: $\sum(a_n-b_n)=\sum a_n-\sum b_n = A-B$

• Constant Multiple rule: $\sum(ka_n)=k\sum a_n= kA$

Corollary

• If $\sum a_n$ diverges, then $k \sum a_n$ for $k\neq 0$?

• If $\sum a_n$ diverges and $\sum b_n$ converges, then $\sum(a_n+b_n)$ ?

Example: $\sum\limits_{n=1}^\infty {3^{n-1}-1\over 6^{n-1}}$

Adding or Deleting Terms

Adding or deleting a finite number of terms does not change the series's convergence or divergence.

Reindexing

The following four notations are equivalent. $$\sum_{n=1}^\infty {1\over 2^{n-1}},\qquad \sum_{n=0}^\infty{1\over 2^n},\qquad \sum_{n=5}^\infty {1\over 2^{n-5}},\qquad \sum_{n=-4}^\infty {1\over 2^{n+4}}$$

Nondecreasing Partial Sums

Consider $\sum\limits_{n=1}^\infty a_n$ with $a_n\geq 0$ for all $n$. We have

Corollary

A series $\sum\limits_{n=1}^\infty a_n$ of nonnegative terms converges if and only if its partial sums $\{s_n\}$ are bounded from above.

• Does the harmonic series $\sum\limits_{n=1}^\infty {1\over n}$ converge or not? (Check the Comparison Test)

The Integral Test (Theorem 9--The integral test)

Let $\{a_n\}$ be a sequence of positive terms. Suppose that $a_n=f(n)$, where $f$ is a continuous, positive, decreasing function of $x$ for all $x\geq N$. Then the series $\sum\limits_{n=N}^\infty a_n$ and the integral $\int\limits_N^\infty f(x)dx$ both converge or both diverge.

• $p$-series: $\sum\limits_{n=1}^\infty {1\over n^p}$

Consider $$\int_1^\infty {1\over x^p}dx$$ If $p=1$, we have that $$\int_1^\infty {1\over x^p}dx=\ln (+\infty)=\infty$$ If $p\neq 1$, we have that $$\int_1^\infty {1\over x^p}dx={1\over 1-p}x^{1-p}\big|_{1}^\infty$$ It converges when $p>1$ and diverges when $p<1$.

• $\sum\limits_{n=1}^\infty {1\over n^2+1}$

Consider $$\int_1^\infty {1\over x^2+1}dx$$

• $\sum\limits_{n=1}^\infty {1\over n\ln n}$

Consider $$\int_1^\infty {1\over x\ln x}dx$$

Integral Test: Approximation

Though we can show that the series converges, we can not easily find the total sum $S$. But, we can estimate it with the parital sum $s_n$ with the remainder $$R_n=S-s_n=a_{n+1}+a_{n+2}+a_{n+3}+\cdots$$

Estimate $\sum\limits_{n=1}^\infty {1\over n^2}$ with $s_{10}$

We have that $$s_\infty\approx 1.64493$$

The Comparison Test

Let $\sum a_n$, $\sum c_n$, and $\sum d_n$ be series with nonnegative terms. Suppose that for some integer $N$, we have $$d_n\leq a_n\leq c_n,\qquad\forall n>N.$$ Then

• If $\sum c_n$ converges, then $\sum a_n$ also converges, because its partial sum is increasing and bounded.

• If $\sum d_n$ diverges, then $\sum a_n$ also diverges, because its partial sum is unbounded.

Examples

• $\displaystyle\sum\limits_{n=1}^\infty {5\over 5n-1}$ diverges because $${5\over 5n-1}\geq {1\over n}$$

• $\displaystyle\sum\limits_{n=1}^\infty {1\over n!}$ converges because $${1\over n!}\leq {1\over 2^{n-1}}$$

Harmonic series $\sum{1\over n}$

We can find a lower bound of the series.

The Limit Comparison Test

Suppose that $a_n>0$ and $b_n>0$ for $n\geq N$

• If $\displaystyle\lim\limits_{n\rightarrow\infty}{a_n\over b_n}=c>0$, then both $\sum a_n$ and $\sum b_n$ converge or both diverge. (Note $(0.5 c)\cdot b_n<a_n<(1.5c)\cdot b_n$ for large $n$)

• If $\displaystyle\lim\limits_{n\rightarrow\infty}{a_n\over b_n}=0$ and $\sum b_n$ converges, then $\sum a_n$ converges. (Note $a_n<b_n$ for large $n$)

• If $\displaystyle\lim\limits_{n\rightarrow\infty}{a_n\over b_n}=\infty$ and $\sum b_n$ diverges, then $\sum a_n$ diverges. (Note $a_n>b_n$ for large $n$)

Example

• $\displaystyle\sum {2n+1\over n^2+2n+1}$ $$\lim_{n\rightarrow\infty}{{2n+1\over n^2+2n+1}\over {1\over n}}=2$$

• $\displaystyle\sum {1+n\ln n\over n^2+5}$ $$\lim_{n\rightarrow\infty}{{1+n\ln n\over n^2+5}\over {1\over n}}=\infty$$

• $\displaystyle\sum {\ln n \over n^{1.1}}$ $$\lim_{n\rightarrow\infty}{{\ln n \over n^{1.1}}\over {1\over n^{1.05}}}=0$$

What happens with negative terms?

A series $\sum a_n$ converges absolutely (is absolutely convergent) if $\sum |a_n|$ converges.

A absolutely convergent series converges, i.e., if $\sum |a_n|$ converges, then $\sum a_n$ converges.

Because $a_n = {|a_n|+a_n\over 2} - {|a_n|-a_n\over 2}$, and both $\sum {|a_n|+a_n\over 2}$ and $\sum {|a_n|-a_n\over 2}$ converges.

The Ratio Test (Important)

Suppose that $$\lim_{n\rightarrow \infty}\left|{a_{n+1}\over a_n}\right|=\rho$$

• If $\rho<1$, then the series $\sum a_n$ converges.

Compare with $\left({\rho+1\over 2}\right)^n$

• If $\rho>1$ or $\rho =\infty$, the series diverges

$$\lim_{n\rightarrow \infty} |a_n|\neq 0$$

• If $\rho=1$, the test is inconclusive.

$$\sum{1\over n} \text{ and }\sum{1\over n^2}$$

Examples

• $\displaystyle\sum {(2n)!\over n!n!}$

• $\displaystyle\sum {4^nn!n!\over (2n)!}$

Another example

The Root Test

Suppose that $$\lim_{n\rightarrow \infty}\sqrt[n]{|a_n|}=\rho$$

• If $\rho<1$, then the series converges

Compare with $\left(\rho+1\over 2\right)^n$

• If $\rho>1$ or $\rho=\infty$, then the series diverges

$$\lim_{n\rightarrow \infty} |a_n|\neq 0$$

• If $\rho=1$, then the test is inconclusive.

$$\sum{1\over n} \text{ and }\sum{1\over n^2}$$

Alternating Series

An alternating series has terms that are alternatively positive and negative.

The series $\sum_{n=1}^\infty (-1)^{n+1}u_n$ converges if the following are satisfied

• $u_n\geq u_{n+1}>0$ for $n\geq N$.

• $u_n\rightarrow 0$.

WLOG, assume that $N=1$. Consider $s_{2m}$

• $s_{2m}$ is increasing, while $s_{2m+1}$ is decreasing.

• $0\leq s_{2m}\leq s_{2m+1}\leq u_1$.

• Both $s_{2m}$ and $s_{2m+1}$ converge, and $u_{2m+1}\rightarrow 0$.

Example

• $\sum\limits_{n=1}^\infty (-1)^{n+1}{1\over n}$