Infinite Series
December 8, 2016
Categorised in: 1st Year Maths 1
Infinite series is denoted as:
$$∑↙{n=1}↖∞ u_n$$ or $$∑u_n$$
Sum of first n terms in series is denoted by:
$S_n = u_1 + u_2 + … + u_n$
Convergent series: $S_n$ tends to finite limit S as n tends to infinity.
$S = \lim↙{n→∞} S_n$
Divergent series: $S_n$ tends to +∞ or -∞ as n tends to infinity.
$\lim↙{n→∞} S_n = +∞ or -∞ $
Oscillatory Series: $S_n$ neither tends to a finite limit nor to +∞ or -∞ as n tends to infinity.
General:
* $\lim S_n$ = finite quantity -> series is convergent.
* Series is positive and $\lim u_n > 0$ -> series is divergent.
Test of Convergence and Divergence:
Cauchy’s $n^{th}$ root test:
Convergent: $\lim↙{n→∞} (u_n)^{1/n} < 1$
Divergent: $\lim↙{n→∞} (u_n)^{1/n} > 1$
$\lim↙{n→∞} (u_n)^{1/n} = 1 \text" test fails"$
P series
$$∑↙{n=1}↖{∞} {1}/{n^P}$$
Convergent: p>1
Divergent: p≤1
Comparison test:
$\lim↙{n→∞} {u_n}/{v_n} \text" is finite" $
Here, $v_n = {1}/{n^{q-p}}$ where p and q are respectively the highest indices of n in the numerator and denominator of $u_n$.
Appl P series test on $v_n$
D’Alembert’s ratio test:
Convergent: $\lim↙{n→∞} {u_n}/{u_{n+1}} > 1$
Divergent: $\lim↙{n→∞} {u_n}/{u_{n+1}} < 1$
$\lim↙{n→∞} {u_n}/{u_{n+1}} = 1$ test fails
Auxiliary series $∑{1}/{n(log n)^p}$
Convergent: p > 1
Divergent: p≤1
Gauss’s test:
${u_n}/{u_{n+1}} = 1 + l/n + {b_n}/{n^p}$
p>1 and $b_n$ bounded as n -> ∞
Convergent: $l$>1
Divergent: $l$≤1
Logarithmic test:
Convergent: $\lim↙{n→∞} (n log{u_n}/{u_{n+1}}) > 1$
Divergent: $\lim↙{n→∞} (n log{u_n}/{u_{n+1}}) < 1$
De Morgan’s or Bertrand’s test:
Convergent: $\lim↙{n→∞} [{[n{({u_n}/{u_{n+1}} – 1)} -1]}log n] > 1$
Divergent: $\lim↙{n→∞} [{[n{({u_n}/{u_{n+1}} – 1)} -1]}log n] < 1$
Pratik Kataria is currently learning Springboot and Hibernate.
Technologies known and worked on: C/C++, Java, Python, JavaScript, HTML, CSS, WordPress, Angular, Ionic, MongoDB, SQL and Android.
Softwares known and worked on: Adobe Photoshop, Adobe Illustrator and Adobe After Effects.