Who can deny that life is not fair, nor easy, just a messy affair where we muddle along with glimpses of significant relationships, search of meaning and purpose? Who can justify that lies are abundant and spreading, news are mainly fake and driven by political and economic narratives, 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, money is king and sometimes even god, and life is but a dream, a worn out ship on troubled waters? Who can deny that I love you Bew like crazy, my sweet angel? Desperate verses IV, M. Anawim, #justtothepoint.
A finite series is given by all the terms of a finite sequence, added together. 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.
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] $S_n -rS_n = a - ar^{n} ⇒ S_n(1-r) = a(1-r^n) ⇒$[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] = $\frac{1}{1-\frac{1}{2}} = \frac{1}{1/2} = 2.$
$\lim_{n \to ∞}S_n = \lim_{n \to ∞} (\frac{3}{4}-\frac{1}{2n}-\frac{1}{2(n+1)}) = \frac{3}{4}$. 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 $\sum_{n=1}^\infty \frac{1}{3^{n-1}} = \frac{a}{1-r} = \frac{1}{1-\frac{1}{3}} = \frac{3}{2}.$
$\sum_{n=1}^\infty \frac{1}{n}$. Taking the upper Riemann sum (Δx = 1), $\int_{1}^{n} \frac{dx}{x} < 1 + \frac{1}{2} + \frac{1}{3} + ··· + \frac{1}{N-1}$ <[S_{n} has one more term, namely ^{1}⁄_{N}] S_{N}.
Besides, $\int_{1}^{n} \frac{dx}{x} = ln(x)\bigg|_{1}^{N} = ln(N)$ ⇒ ln(N) < S_{n} (Figure 2.1).
Futhermore, we know that $\lim_{N \to ∞}ln(N) = ∞$, hence $\lim_{N \to ∞}S_N = ∞$ ⇒ $\sum_{n=1}^\infty \frac{1}{n}$ diverges.
Taking the lower Riemann Sum (Δx = 1), $\int_{1}^{n} \frac{dx}{x} > \frac{1}{2} + \frac{1}{3} + ··· + \frac{1}{N} = S_N -1$ ⇒ ln(N) < S_{N} < ln(N) + 1. We can generalize this result in the following proposition.
Necessary condition for the convergence of a series. If $\sum_{n=1}^\infty a_n$ converges, then $\lim_{n \to ∞}a_n = 0$
Proof. (Idea: In order for a series to converge the series terms must go to zero in the limit.)
a_{n} = s_{n} -s_{n-1}. Besides, if $\sum_{n=1}^\infty a_n$ converges ⇒ $\sum_{n=1}^\infty s_n$ is also convergent for some value s ↭ $\lim_{n \to \infty} s_n = s$ ⇒ $\lim_{n \to \infty} s_{n-1} = s$ ⇒ $\lim_{n \to ∞}a_n = \lim_{n \to ∞} s_n - s_{n-1} = \lim_{n \to ∞} s_n - \lim_{n \to ∞} s_{n-1} = s -s = 0$.
Example. Be cautious!💀 This result gives us a necessary condition or requirement for convergence but it is not a guarantee of convergence. The series terms of $\sum_{n=1}^\infty \frac{1}{n}$ and $\sum_{n=1}^\infty \frac{1}{n^2}$ are zero in the limit as n goes to infinity, yet only the second series converges.
Integral Comparison. If f(x) is a positive, continuous, and decreasing function on the interval [1, ∞), then $|\sum_{n=1}^\infty f(n) -\int_{1}^{∞} f(x)dx| < f(1).$ Besides, $\sum_{n=1}^\infty f(n)$ converges, if and only if, $\int_{1}^{∞} f(x)dx$ converges.
Proof.
Using the same argument (upper Riemann sum, the rectangles enclose more area than the area under f(x)), $\int_{1}^{∞} f(x)dx < \sum_{n=1}^\infty f(n)$ (Figure 2.a), and $\sum_{n=2}^\infty f(n) < \int_{1}^{∞} f(x)dx$ (Figure 2.b., the areas of the rectangles is less than the area under f(x), but this summation starts with n = 2) ⇒ $\sum_{n=1}^\infty f(n) < a_1 + \int_{1}^{∞} f(x)dx$ where a_{1} = f(1).
Combining both equations, $\sum_{n=1}^\infty f(n) < a_1 + \int_{1}^{∞} f(x)dx < a_1 + \sum_{n=1}^\infty f(n)$ ⇒ $|\sum_{n=1}^\infty f(n) -\int_{1}^{∞} f(x)dx| < f(1).$
Notice that $\sum_{n=1}^\infty f(n) < a_1 + \int_{1}^{∞} f(x)dx < a_1 ⇒ \sum_{n=1}^\infty f(n) - \int_{1}^{∞} f(x)dx < a_1$. Besides, $a_1 + \int_{1}^{∞} f(x)dx < a_1 + \sum_{n=1}^\infty f(n) ⇒ \sum_{n=1}^\infty f(n) - \int_{1}^{∞} f(x)dx > 0 ≥ -a_1,$ because f is positive f(1) = a_{1}≥ 0.
If $\sum_{n=1}^\infty f(n)$ diverges, so does $\int_{1}^{∞} f(x)dx$ because $\sum_{n=1}^\infty f(n) < a_1 + \int_{1}^{∞} f(x)dx$.
If $\sum_{n=1}^\infty f(n)$ converges, so does $\int_{1}^{∞} f(x)dx$ because $\int_{1}^{∞} f(x)dx < \sum_{n=1}^\infty f(n)$.
Check conditions: f(x) = $\frac{ln(n)}{n^2}$ is positive, continuos, and decreasing on the interval [2, ∞). First, we know that f(x) is positive and continuous as both ln(n) and n^{2} are positive and continuous. f’(x) = $\frac{x-2x·ln(x)}{x^4} = \frac{1-2ln(x)}{x^3}$.
x ≥ 2 ⇒[We indeed only need x ≥ $e^{\frac{1}{2}}$ ≈ 1.65] ln(x) ≥ ^{1}⁄_{2} ⇒ 2ln(x) ≥ 1 ⇒ 1 -2ln(x) ≤ 0 ⇒ ∀x ≥ 2, f’(x) ≤ 0 ⇒ ∀x ≥ 2, f is decreasing.
Integral test: $\int_{1}^{∞} \frac{ln(x)}{x^2}dx = \lim_{a \to ∞} \int_{1}^{a} \frac{ln(x)}{x^2}dx =$ [Integration by parts, u = ln(x), dv = ^{dx}⁄_{x2}, du = ^{dx}⁄_{x}, v = ^{-1}⁄_{x}] =[1]
$\lim_{a \to ∞} -\frac{ln(x)}{x}\bigg|_{1}^{a}$ =
$\lim_{a \to ∞} -\frac{ln(a)}{a} +ln(1) = \lim_{a \to ∞} -\frac{ln(a)}{a}$ [Let’s apply L’Ho^pital’s Rule] $\lim_{a \to ∞} -\frac{\frac{1}{a}}{1} = \lim_{a \to ∞} -\frac{1}{a} = 0$
$\lim_{a \to ∞} \int_{1}^{a} \frac{1}{x^2} dx = \lim_{a \to ∞}-\frac{1}{x}\bigg|_{1}^{a} = \frac{-1}{a}+1$ → 1 when a → ∞. Therefore, [1] = 0 + 1 = 1, since the integral converges, the series converges (the first term f(1) is just a constant).
Theorem. Direct Comparison test. Let {a_{n}} and {b_{n}} be positive sequences where a_{n}≤b_{n} ∀n≥N, for some N.
In this case, let’s compare the given series to $\sum_{n=1}^\infty \frac{1}{3^n}$. This is a convergent geometric series, it converges because ^{1}⁄_{3}< 1.
Since $3^n < 3^n + n^2 ⇒ \frac{1}{3^n} > \frac{1}{3^n+n^2}$. The series $\sum_{n=1}^\infty \frac{1}{3^n}$ is a convergent one, then our original series converges, too.
Since ∀n ≥ 1, n ≥ n -ln(n), $\frac{1}{n} ≤ \frac{1}{n-ln(n)}$, and whe know that the harmonic series diverges $\sum_{n=1}^\infty \frac{1}{n}$, so we can conclude that $\sum_{n=1}^\infty \frac{1}{n-ln(n)}$ diverges.
Theorem. If $\lim_{n \to ∞} \frac{f(n)}{g(n)} = 1$ (f(n) ~ g(n) as n → ∞), g is positive, then $\sum_{n=1}^\infty f(n), \sum_{n=1}^\infty g(n)$ either both converge or diverge.
Example. $\sum_{n=1}^\infty \frac{1}{\sqrt{n^2+1}}$ diverges because $\sum_{n=1}^\infty \frac{1}{\sqrt{n^2}} = \sum_{n=1}^\infty \frac{1}{n}$ diverges.
Rules: No glue! Blocks have to be placed and supported entirely by their own weights. Only one block per level. We’re making a skewed tower. All blocks are of the same shape, weight, and are of uniform density.
The best strategy is a “top-down” approach, that is, to build from the top block down. Let C_{0} and C_{1} be the left end and the center of mass of the first block (top block) respectively, C_{0} = 0, C_{1} = 1. Put the second block as far to the right as possible so that it’s left end is at C1.
Let C2 be the center of mass of the top two blocks (C_{2} = 1 + ^{1}⁄_{2}). Once again put the left end of the next block underneath the center of mass of all the previous blocks combined (C_{3} = 1 + ^{1}⁄_{2} + ^{1}⁄_{3}), and in general, $C_{n+1} = \frac{nC_n +1(C_n + 1)}{n + 1}$ [the strategy is to put the left end of the next block underneath the center of mass of all the previous ones combined, namely C_{n} + 1] = $\frac{(n+1)C_n+1}{n+1} =C_n + \frac{1}{n+1}$.
Recall the Riemann Sum estimation from a previous exercise, ln(N) < S_{N} < ln(N) + 1 where S_{N} = C_{N} = 1 + ^{1}⁄_{2} + ^{1}⁄_{3} + ··· + ^{1}⁄_{N}. Sol: You can extend this stack of blocks as far as you want provided that you have enough blocks because ln(N) → ∞ ⇒ S_{N} → ∞
As long as we are strictly inside the interval of convergence, we can take derivatives and integrals of power series allowing us to get new series.
Let R be the radius of convergence of a series, x is strictly inside the interval of convergence of the series when −R < x < R.
To differentiate, we simply differentiate each term (not worrying that we have infinitely many terms) and then put the terms back into summation notation. Notice that in the derivative series we must change our index to begin at n = 1.
$\frac{d}{dx}(\sum_{n=0}^\infty a_nx^n) = \frac{d}{dx}(\sum_{n=1}^\infty a_0 + a_1x +a_2x^2 +a_3x^3 +···) = a_1 + 2a_2x + 3a_3x^2 +··· = \sum_{n=1}^\infty na_nx^{n-1}$
Similarly, while we get an infinite integrand, we don’t worry and just antidifferentiate each term, and then add a constant.
$\int \sum_{n=0}^\infty a_nx^n = \int a_0 + a_1x +a_2x^2 +a_3x^3 + ··· = c + a_0x + a_1\frac{x^2}{2} + a_2\frac{x^3}{3} + ··· = (\sum_{n=0}^\infty a_n \frac{x^{n+1}}{n+1}) + C$