Who can deny that life is not fair, nor easy, just a messy affair where we muddle along with glimpses of significant relationships and futile searches of meaning and purpose? Who can justify that lies are abundant and spreading, news are mainly propaganda driven by political and economic narratives, authorities are corrupt, history is being twisted, trust is all gone, and we all know that something needs to be done? Who can deny that yesterday is gone, tomorrow is not promised, and we only have today?
Who can rebuff that destiny is a bitch, the world is getting crazy, money is king and sometimes even god, and life is but a dream, a worn out ship on troubled waters, and things are only getting harder? Who can deny that I love you Bew like crazy, my sweet angel? Desperate verses IV, M. Anawim, #justtothepoint.
A sequence is a list of things (usually numbers) arranged in a definite order according to some rule or pattern. Each number in the sequence is called a term, e.g., {3, 5, 7, . . . , 21}, {3, 2, 1} is 3 to 1 backwards, {a, b, c, …, z} is the sequence of the alphabetic letters, and {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} is the sequence of the first 10 prime numbers.
The rational number $\frac{3}{9}$ can be written as 0.33333333333… = $0.\widehat{3}$ and as an infinite series as follows 0.3 + 0.03 + 0.003 + ···.
A finite series is given by all the terms of a finite sequence, added together, e.g., {3, 5, 7, . . . , 21}, $\sum_{k=1}^{10} 2k+1 = 120$. An infinite series is the sum of an infinite sequence of numbers. It is represented in the form $\sum_{n=1}^\infty a_n = a_1 + a_2 + a_3 + ···$ where a_{n} represents the terms of the sequence, and n is the index that ranges from 1 to infinite.
A series is convergent (or converges) if the sequence of its partial sums tends to a limit, that is, l = $\lim_{n \to ∞} \sum_{k=1}^n a_k$ exists and is a finite number. More precisely, if there exists a number l (or S) such that for every arbitrary small positive number ε, there is a (sufficient large) N, such that ∀n ≥ N, |S_{n} -l| < ε where S_{n} = $\sum_{k=1}^n a_k = a_1 + a_2 + ··· + a_n$. If the series is convergent, the number l is called the sum of the series. On the contrary, any series that is not convergent ($\lim_{n \to ∞} \sum_{k=1}^n a_k$ does not exist) is said to be divergent or to diverge.
An arithmetic series is the sum of a sequence in which each term is computed from the previous one by adding (or subtracting) a constant. ∀ k≥ 1, a_{k} = a_{k-1} + d = a_{k-2} + 2d = ··· = a_{1} + d(k-1).
The nth partial sum of an arithmetic sequence is] $S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n a_1 +(k-1)d = n·a_1+d\sum_{k=1}^n (k-1) = n·a_1+d\sum_{k=2}^n (k-1) = n·a_1+d\sum_{k=1}^{n-1} k $[Sum identity, $\sum_{k=1}^n k = \frac{1}{2}n(n+1)$] = $n·a_1+d·\frac{1}{2}(n-1)n = \frac{1}{2}n(2a_1+d(n-1))$⇒[$a_1+a_n = a_1 + a_1+d(n-1)=2a_1+d(n-1)$] $S_n = \frac{n·(a_1+a_n)}{2}$.
$S_n = n·a_1+d·\frac{1}{2}(n-1)n = (a_1-\frac{d}{2})n+\frac{d}{2}n^2$. Since both quadratic and linear functions tend to infinity as n increases, the only way that an arithmetic series converges is when $\frac{d}{2}$ and $a-\frac{d}{2}$ are both equal to zero ⇒ a = d = 0, so the only arithmetic series that converges is 0 + 0 + 0 + ···
A geometric series is a specific type of infinite series where each term is obtained by multiplying the previous term by a fixed, non-zero constant. The general form of a geometric series is: $\sum_{n=0}^\infty a_n = \sum_{n=0}^\infty ar^n = a + ar + ar^2 + ar^3 + ···$
$S_n = a + ar + ar^2 + ar^3 + ··· + ar^{n-1}$ ⇒[Multiplying both sides of the equation by r] $rS_n = ar + ar^2 + ar^3 + ar^4 + ··· + ar^{n}$ ⇒[Subtracting these equations we then obtain] $S_n -rS_n = a - ar^{n} ⇒ S_n(1-r) = a(1-r^n) ⇒$[Assume r ≠ 1, we can divide both sides by 1-r and obtain the formula for the nth partial sum of a geometric sequence] $S_n = \frac{a(1-r^n)}{1-r}$
If |r| < 1, $\lim_{n \to ∞}S_n = \lim_{n \to ∞} \frac{a(1-r^n)}{1-r} = \frac{a(1-0)}{1-r} = \frac{a}{1-r}$. Otherwise, |r| > 1, the series diverges.
The geometric series converges if and only if the common ratio |r| is strictly less than 1, and the sum S of a convergent geometric series is given by the formula S = $\frac{a}{1-r}$, e.g., $\sum_{n=0}^\infty \frac{1}{2^n} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ···$= [This is a geometric series with a = 1 and r = 1/2, |r| < 1] = $\frac{1}{1-\frac{1}{2}} = \frac{1}{1/2} = 2.$
Let $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ be convergent series. Then, the following algebraic properties hold (Credits: Monroe_Community_College, Mathematics Libre Texts):
$\frac{3}{9} = 0.3 + 0.03 + 0.003 + ··· = \frac{3}{10} + \frac{3}{10^2} + \frac{3}{10^3} + ···. \sum_{n=1}^\infty \frac{3}{10^n}$ this is a geometrical serie where a = 3/10, and r = $\frac{1}{10}< 1$, hence $\sum_{n=1}^\infty \frac{3}{10^n} = \frac{\frac{3}{10}}{1-\frac{1}{10}} = \frac{\frac{3}{10}}{\frac{9}{10}} = \frac{3}{9}.$
1 + 1 + 1 + 1 + ··· 1 = $\sum_{n=1}^\infty 1 = ∞$. It is a divergent series because the sum of the terms does not approach a finite value, $\lim_{n \to ∞} S_n = \lim_{n \to ∞} n = ∞$.
$\sum_{n=1}^\infty \frac{(-3)^{n+1}}{4^{n-1}}$ this is a geometrical serie where a = $\frac{(-3)^{2}}{4^{0}}=9$, and r = $\frac{-3}{4}< 1$, hence $\sum_{n=1}^\infty \frac{(-3)^{n+1}}{4^{n-1}} = \frac{9}{1+\frac{3}{4}}=\frac{36}{7}.$
$\sum_{n=1}^\infty \frac{1}{n(n+1)} = \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+···$
The general formula for the partial sum is $s_n = \sum_{k=1}^n \frac{1}{k(k+1)}$ =[We can use partial fraction decomposition to express it as] $\sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{k=1}^n \frac{k+1-k}{k(k+1)} = \sum_{k=1}^n \frac{k+1}{k(k+1)}-\frac{k}{k(k+1)} = \sum_{k=1}^n \frac{1}{k}-\frac{1}{(k+1)} = (\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3}) + (\frac{1}{3}-\frac{1}{4}) + ··· + (\frac{1}{n-1}-\frac{1}{n}) + (\frac{1}{n}-\frac{1}{n+1})$[As the reader could easily see, most of the terms in this expression cancel in pairs.]= $1 -\frac{1}{n+1} = \frac{n+1-1}{n+1} = \frac{n}{n+1}$
$\lim_{n \to ∞}S_n = \lim_{n \to ∞} \frac{n}{n+1} = 1$. The sequence of partial sums converges and its value is $\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1$.
$\lim_{n \to ∞}S_n = \lim_{n \to ∞} \frac{1}{2}-\frac{1}{n+2} = \frac{1}{2}$. The sequence of partial sums converges and its value is $\sum_{n=2}^\infty (\frac{1}{n+1}-\frac{1}{n+2}) = \frac{1}{2}$
$\lim_{n \to ∞}S_n = \lim_{n \to ∞} (\frac{3}{4}-\frac{1}{2n}-\frac{1}{2(n+1)}) = \frac{3}{4}$. The sequence of partial sums converges and its value is $\sum_{n=2}^\infty \frac{1}{n^2-1} = \frac{3}{4}$
1 - 1 - 1 - 1 + ···, S_{n} = 1 −1 +1 −1 + ···+ (−1)^{n}. Depending on whether n is even or odd, the partial sum alternates between 0 and 1. Formally, for even n, S_{n} = 0, and for odd n, S_{n} = 1 ⇒ S_{n} does not converge to a fixed value as n approaches infinity, so the series 1 - 1 - 1 - 1 + ··· is divergent.
$\sum_{n=1}^\infty \frac{1}{3^{n-1}} = 1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3}+···$. It is a geometric series where a = 1 (the first term), r = 1/3, |r|< 1, so the series is convergent and its sum is $\sum_{n=1}^\infty \frac{1}{3^{n-1}} = \frac{a}{1-r} = \frac{1}{1-\frac{1}{3}} = \frac{3}{2}.$
$\sum_{n=1}^\infty (\frac{3}{n(n+1)}+\frac{1}{3^{n-1}})$ is convergent because we have previously demonstrated that $\sum_{n=1}^\infty \frac{1}{3^{n-1}} = \frac{3}{2},$ and $\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1$ ⇒ By using the Algebraic Properties of Convergent Series, $\sum_{n=1}^\infty (\frac{3}{n(n+1)}+\frac{1}{3^{n-1}}) = 3·\sum_{n=1}^\infty (\frac{1}{n(n+1)})+ \sum_{n=1}^\infty \frac{1}{3^{n-1}} = 3·1 + \frac{3}{2} = \frac{9}{2}$.