Module 8 Formulas: Series and Sequences
Sequence
Sequences: A sequence is a list of numbers, or terms, arranged in an orderly fashion. They are usually denoted by curly brackets, \{\} and written either explicitly as \{a_1,a_2,\ldots,a_n,\ldots\} or \{a_n\}. The subscript n is called the sequence’s index and can be any variable; n is commonly used, but i or k can also be used instead. Some examples include
\begin{aligned} \left\{\frac 1{n^2}\right\}_{n=1}^{\infty} & =\left\{1,\frac 14,\frac 19,\frac 1{16},\ldots,\frac 1{n^2},\ldots\right\} \\ \left\{\frac n{2n+1}\right\}_{k=0}^{N} & =\left\{0,\frac 13,\frac 25,\frac 37,\ldots,\frac N{2N+1}\right\} \end{aligned}
Notice how the second sequence starts from k=0 and ends at k=N. This sequence is finite whereas the first sequence starts from n=1 and goes indefinitely up until k\to+\infty; this sequence is infinite. In fact, sequences have a certain behavior as n\to+\infty: they may either converge or diverge, like any other function! For a sequence \{a_n\} to converge, \lim\limits_{n\to+\infty}a_n=L must exist and be a single number. If a single value for L does not exist, then the sequence is said to diverge.
Squeeze Theorem: If b_n\leq a_n\leq c_n for some n\geq n_0, and L=\lim\limits_{n\to+\infty}b_n=\lim\limits_{n\to+\infty}c_n, then
\lim\limits_{n\to+\infty}a_n=L \tag{1}
For the squeeze theorem to apply, b_n and c_n must both converge to the same L value as n\to+\infty, otherwise the final value of a_n as n\to+\infty is indeterminant.
Certain sequences will contain a function known as a “factorial”. These are special functions that can be recursively defined using itself. For a factorial n!, then the definition is
n!=\prod\limits_{k=1}^nk=n\times(n-1)\times(n-2)\times(n-3)\times\ldots\times2\times1 \tag{2}
Eq. 2 is recursive since we can notice that the boxed portion below is another instance of the factorial function!
n!=n\times\boxed{(n-1)\times(n-2)\times(n-3)\times\ldots\times2\times1}=n\times(n-1)! \tag{3}
Notice how (\cdot)! appears on the left and right-hand side of Eq. 3. This is what “recursive” is describing above. This definition allows us to simplify terms involving factorials like so.
\frac {k!}{(k+2)!}=\frac {k!}{(k+2)(k+1)\color{blue}{k(k-1)(k-2)\ldots1}}=\frac {k!}{(k+2)(k+1)\color{blue}{k!}}=\frac 1{(k+1)(k+2)}
Series
Geometric Series: The radius of convergence is |r|<1 for Eq. 4 to converge. In other words, if |r|\geq1, then Eq. 4 will diverge. \sum\limits_{n=1}^{+\infty}ar^{n-1}=\frac a{1-r} \tag{4}
Divergence Test: For a series \sum a_n, if L is defined as
L=\lim\limits_{n\to+\infty}a_n
then if |L|>0, then the series is automatically divergent. If L=0, the test is inconclusive and the series may be either convergent or divergent.
Convergence Tests
Ratio Test: If \sum\left|a_n\right| is convergent, then \sum a_n is absolutely convergent. If L is defined as
L=\lim\limits_{n\to+\infty}\left|\frac {a_{n+1}}{a_n}\right|
then L<1 means the series is absolutely convergent and therefore, convergent. If L>1, then the series is divergent. And if L=1, the case is inconclusive and a different convergence test will be needed.
Power Series
Power Series: A series of the form below is referred to as a power series
\sum\limits_{n=0}^{+\infty}c_n(x-a)^n=c_0+c_1(x-a)+c_2(x-a)^2+\ldots \tag{5}
where c_n is the power series’ coefficients and a is the center.
Radius of Convergence: Unfortunately, not all values of x may lead to a power series converging. The interval containing all possible values of x that results in the power series converging is called the interval of convergence and is written as |x-a|<R. The value for R is called the radius of convergence. This interval describes a region that is centered on a, with either endpoints spaced a distance R from a. Therefore, the endpoints for the interval above are (a-R, a+R).
If the series converges for all x, then R=\infty (the radius of the interval a distance from the center a) and the physical interval is (-\infty, +\infty).
If the series converges for some values of x, then the radius R is a real number such that |x-a|<R results in a converging series and |x-a|>R results in a diverging series.
If the series converges for a single value of x, then R=0 and the physical interval only contains the center \{a\}.
If the interval of convergence is unknown for a brand new series, then apply the Ratio Test to derive the interval. That is, write L=\lim\limits_{n\to+\infty}\left|\frac {a_{n+1}}{a_n}\right|<1 and evaluate the limit to rewrite into the form |x-a|<R.
You may have noticed that both intervals of convergence shown in point 2 only show a < or > sign, not a \leq or \geq sign. This is because if the interval of convergence is being created, whether or not the endpoints are included must be determined by substituting in the value for x at the endpoint and testing whether the series converges. That is, if the interval of convergence is currently |x-a|<R, then whether the endpoints x=a-R and x=a+R should be included in the interval can only be determined by substituting x=a\pm R and using any convergence test to determine if the series converges.
Again, you only need to do this if you are unsure of the radius or interval of convergence. If the radius or convergence interval is already known, then plugging in the endpoints is not needed.
Functions as Power Series
Geometric Series: As mentioned in Sec. 2, if |x|<1 and a=1, then
\frac 1{1-x}=\sum\limits_{n=0}^{+\infty}x^n
This in and of itself is sneakily a power series centered on x=0 with constant coefficients c_n=1. The radius of convergence is R=1 since |x|<1.
Natural Logarithm: If |x|<1, then
\ln(1+x)=\sum\limits_{n=1}^{+\infty}(-1)^{n-1}\frac {x^n}n=x-\frac {x^2}2+\frac {x^3}3-\frac {x^4}4+\ldots \tag{6}
Arctangent: If |x|<1, then
\arctan x=\sum\limits_{n=0}^{+\infty}(-1)^n\frac {x^{2n+1}}{2n+1}=x-\frac {x^3}3+\frac {x^5}5-\frac {x^7}7+\frac {x^9}9-\ldots \tag{7}
Taylor and Maclaurin Series
Taylor Series: A special case of the power series occurs when the coefficients can be calculated based off the function’s nth derivative. For a function f(x) centered on x=a, it’s Taylor Series expansion is
f(x)=\sum\limits_{n=0}^{+\infty}\frac {f^{(n)}(a)}{n!}(x-a)^n=f(a)+\frac {f'(a)}{1!}(x-a)+\frac {f''(a)'}{2!}(x-a)^2+\ldots+\frac {f^{(n)}(a)}{n!}(x-a)^n+\ldots \tag{8}
Taylor Polynomial: If Eq. 8 is truncated at some finite point, then the resulting sum is called the Taylor polynomial. For instance, a 3rd order Taylor polynomial for Eq. 8 is
T_3(x)=f(a)+\frac {f'(a)}{1!}(x-a)+\frac {f''(a)}{2!}(x-a)^2+\frac {f'''(a)}{3!}(x-a)^3
Maclaurin Series: A special case of the Taylor Series, when a=0, then
f(x)=\sum\limits_{n=0}^{+\infty}\frac {f^{(n)}(0)}{n!}x^n=f(0)+\frac {f'(0)}{1!}x+\frac {f''(0)'}{2!}x^2+\ldots+\frac {f^{(n)}(0)}{n!}x^n+\ldots \tag{9}
Exponential Function Maclaurin Series: The radius of convergence is R=+\infty (this series converges for any x).
e^x=\sum\limits_{n=0}^{+\infty}\frac {x^n}{n!}=1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+\ldots+\frac {x^n}{n!}+\ldots \tag{10}
Sine Function Maclaurin Series: The radius of convergence is R=+\infty (this series converges for any x).
\sin x=\sum\limits_{n=0}^{+\infty}\frac {(-1)^n}{(2n+1)!}x^{2n+1}=x-\frac {x^3}{3!}+\frac {x^5}{5!}-\frac {x^7}{7!}+\ldots+(-1)^n\frac {x^{2n+1}}{(2n+1)!}+\ldots \tag{11}
Cosine Function Maclaurin Series: The radius of convergence is R=+\infty (this series converges for any x).
\cos x=\sum\limits_{n=0}^{+\infty}\frac {(-1)^n}{(2n)!}x^{2n}=1-\frac {x^2}{2!}+\frac {x^4}{4!}-\frac {x^6}{6!}+\ldots+(-1)^n\frac {x^{2n}}{(2n)!}+\ldots \tag{12}
Binomial Series: If |x|<1 (i.e., the radius of convergence is R=1), then
(1+x)^k=\sum\limits_{n=0}^{+\infty}\binom knx^n=\binom k0+\binom k1x+\binom k2x^2+\ldots=1+kx+\frac {k(k-1)}{2!}x^2+\frac {k(k-1)(k-2)}{3!}x^3+\ldots \tag{13}
Binomial Coefficient: The coefficient \binom kn in Eq. 13 is called the binomial coefficient and is defined as
\binom kn=\frac {k!}{n!(k-n)!}=\frac {k(k-1)(k-2)\ldots(k-n+1)}{n!}=\binom k{k-n} \tag{14}
A shortcut for evaluating Eq. 14 is to start with k and successively subtract 1 from k until a total of k factors are obtained, then divide by n!. For instance, with
\binom 74=\frac {\overbrace{7\cdot6\cdot5\cdot4}^{\text{4 copies}}}{4!}=\frac {840}{24}=35