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The word ‘politics’ is derived from the word ‘poly’, meaning ‘many’, and the word ‘ticks’, meaning ‘blood-sucking parasites’, Larry Hardiman.
An infinite series is the sum of an infinite sequence of numbers. It is represented as: $\sum_{n=1}^\infty a_n = a_1 + a_2 + a_3 + \dots$ 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 finite limit, that is, $l = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \sum_{k=1}^n a_k$ exists and is a finite number.
The $n$-th partial sum $S_n$ is the sum of the first $n$ terms: $S_n = \sum_{k=1}^n a_k = a_1 + a_2 + \dots + a_n$. If the limit $L = \lim_{n \to \infty} S_n$ exists and is a finite number, the series converges, and $L$ is called the sum of the series. Formally, this means for any arbitrarily small positive number $\epsilon$, there is a sufficiently large $N$ such that for all $n \geq N$, $\vert{}S_n - L\vert{} < \epsilon$.
If this limit does not exist (e.g., it goes to infinity or oscillates forever), the series is said to be divergent.
The Divergence Test (n-th Term Test). If $\lim_{n \to \infty} a_n \neq 0$, then the series $\sum_{n=1}^\infty a_n$ diverges.
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.
When dealing with a complex series, it is often easier to compare it to a simpler “benchmark” series whose convergence behavior is already known. The two most important families of benchmark series are geometric series and p-series.
Theorem (p-Series): A p-series $\sum_{n=1}^\infty \frac{1}{n^p}$ converges if and only if $p > 1$.
Proof:
If $p \leq 1$: We can compare it to the harmonic series. Since $n^p \leq n$ for $p \leq 1$, taking the reciprocal gives $\frac{1}{n^p} \geq \frac{1}{n}$. Because the harmonic series $\sum \frac{1}{n}$ diverges, any series whose terms are larger must also diverge.
If $p > 1$: The function $f(x) = \frac{1}{x^p} = x^{-p}$ is positive, continuous, and decreasing for $x \geq 1$. We can use the Integral Test:
$\int_{1}^{\infty} x^{-p} dx = \lim_{b \to \infty} \left[ \frac{x^{-p+1}}{-p+1} \right]_{1}^{b} = \lim_{b \to \infty} \left( \frac{1}{(1-p)b^{p-1}} - \frac{1}{1-p} \right)$
Because $p > 1$, the exponent $(p-1)$ is positive, meaning $b^{p-1} \to \infty$ as $b \to \infty$. Thus, the first term goes to zero: $\int_{1}^{\infty} x^{-p} dx = 0 - \frac{1}{1-p} = \frac{1}{p-1}$
Since the integral evaluates to a finite number, the series converges.
The Direct Comparison Test relies on establishing a strict inequality between the terms of two series.
Theorem (Direct Comparison Test): Let $\sum a_n$ and $\sum b_n$ be series with positive terms such that $0 \leq a_n \leq b_n$ for all $n \geq N$.
Proof.
📐 Partial sums are increasing.
Since $a_n \ge 0, s_{n+1} = s_n + a_{n+1} \ge s_n$, so $\{ s_n \}$ is increasing.
Similarly, $\{ t_n \}$ is increasing.
📐 Step 2 — Comparison of partial sums.
From $a_n \leq b_n$, summing term‑by‑term gives: $s_n = \sum_{i=1}^n a_i \leq \sum_{i=1}^n b_i = t_n$.
Thus, $s_n \leq t_n, \forall n$.
✅ Step 3. If the “bigger” series $\sum b_n$ converges, then the “smaller” series $\sum a_n$ converges.
Then, $\{ t_n \}$ is bounded above, $t_n \leq \sum_{i=1}^\infty b_i$.
Since $s_n \leq t_n, \forall n$, we also have $s_n \leq \sum_{i=1}^\infty b_i$.
Thus, $\{ s_n \}$ is increasing and bounded above. Hence, $\{ a_n \}$ converges.
The Monotone Convergence Theorem states that if a sequence of real numbers is monotonic and bounded, then it must converge.
✅ Step 4. If the “smaller” series $\sum a_n$ diverges, then the “bigger” series $\sum b_n$ diverges.
Since $a_n \ge 0$, divergence means: $s_n \to \infty$.
However, $s_n \leq t_n, \forall n$. Therefore, $t_n \to \infty$.
Thus, $\{ b_n \}$ diverges.
$\sum_{n=1}^\infty \frac{2}{n^3+4} =[\text{We can pull out the constant: }] 2\sum_{n=1}^\infty \frac{1}{n^3+4}$
Besides, $\forall n \ge 1, \frac{1}{n^3+4} \lt \frac{1}{n^3}$ (adding to the denominator makes a fraction smaller), and by the Theorem (p-series) $\sum_{n=1}^\infty \frac{1}{n^3}$ converges (p = 3 > 1) ⇒ By the Direct Comparison Test, $\sum_{n=1}^\infty \frac{1}{n^3+4}$ converges, then the “bigger” series $\sum_{n=1}^\infty \frac{2}{n^3+4}$ converges.
Let $a_n = \frac{1 + \sin(n)}{10^n}$.
The sine function oscillates: $-1 \leq \sin(n) \leq 1$. Adding 1 to all sides: $0 \leq 1 + \sin(n) \leq 2$
Divide by $10^n$: $\frac{0}{10^n} \leq \frac{1 + \sin(n)}{10^n} \leq \frac{2}{10^n} \implies 0 \leq \frac{1+\sin(n)}{10^n} \leq \frac{2}{10^n}$.
Let $b_n = \frac{2}{10^n},~ \sum_{n=0}^\infty \frac{2}{10^n} = 2·\sum_{n=0}^\infty \frac{1}{10^n}=2·\sum_{n=0}^\infty (\frac{1}{10})^n$. This is a convergent geometric series with $r = \frac{1}{10} \lt 1$. Since the “bigger” series converges, our original series converges by the DCT.
Because $\ln(n) > 0$ for $n > 1$, subtracting it makes the denominator smaller, meaning the overall fraction is larger: $n - \ln(n) < n \implies \frac{1}{n - \ln(n)} > \frac{1}{n}$
The series $\sum \frac{1}{n}$ is the divergent harmonic series. Since the “smaller” series diverges, our original series diverges by the DCT.
Often, establishing a strict inequality for the DCT is difficult (e.g., $\frac{1}{n^3-4}$ is larger than $\frac{1}{n^3}$, so comparing it to a convergent series fails the DCT conditions). The Limit Comparison Test solves this by looking only at the growth rates of the dominant terms.
Theorem. Limit Comparison Test. Let $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ be series with positive terms. If $\lim_{n \to \infty}\frac{a_n}{b_n}=c,$ where $c$ is a finite number and $c > 0$, then both series share the same fate: either they both converge or both diverge.
The dominant term in the denominator is $n^3$. Let’s compare our series $a_n$ to the benchmark series $b_n = \frac{1}{n^3}$.
$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{2}{n^3-4}}{\frac{1}{n^3}} = \lim_{n \to \infty} \frac{2n^3}{n^3-4} = 2$
Since $2 > 0$ and finite, the LCT applies. Because $\sum \frac{1}{n^3}$ converges (p-series, $p=3$), our original series converges.
Identify the dominant terms. In the numerator, it’s $\sqrt{n} = n^{1/2}$. In the denominator, it’s $n^2$.
Our benchmark series is $b_n = \frac{n^{1/2}}{n^2} = \frac{1}{n^{3/2}}$. We know $\sum \frac{1}{n^{3/2}}$ converges (p-series, $p=1.5 > 1$).
Now, apply the limit: $\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \left( \frac{\sqrt{n+2}}{2n^2+n+1} \cdot \frac{n^{3/2}}{1} \right) = \lim_{n \to \infty} \frac{\sqrt{n^4+2n^3}}{2n^2+n+1}$
Looking at the highest powers, the numerator is $\sqrt{n^4} = n^2$, and the denominator is $2n^2$. $\lim_{n \to \infty}\frac{\sqrt{n^4+2n^3}}{2n^2+n+1} = \frac{1}{2}$. Analogously, $\lim_{n \to \infty} \frac{\frac{\sqrt{n+2}}{2n^2+n+1}}{\frac{\sqrt{n}}{n^2}} = \lim_{n \to \infty} \frac{\sqrt{n+2}}{2n^2+n+1}·\frac{n^2}{\sqrt{n}} = \lim_{n \to \infty} \sqrt{\frac{n+2}{n}}·\frac{n^2}{2n^2+n+1} = \lim_{n \to \infty} \sqrt{\frac{n+2}{n}}·\lim_{n \to ∞} \frac{n^2}{2n^2+n+1} = 1·\frac{1}{2}=\frac{1}{2} > 0$
Since $\frac{1}{2} > 0$ and finite, our series converges by the LCT.
The dominant terms are the exponentials. Let $b_n = \frac{9^n}{10^n} = \left(\frac{9}{10}\right)^n$. This is obviously a convergent geometric series.
$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \left( \frac{9^n}{3+10^n} \cdot \frac{10^n}{9^n} \right) = \lim_{n \to \infty} \frac{10^n}{3+10^n} = 1$
Since $1 > 0$ and finite, our series $\sum_{n=1}^\infty \frac{9^n}{3+10^n}$ also converges by the LCT.
The dominant term in the numerator is $n 3^n$. The dominant term in the denominator is $n^3$. Let $b_n = \frac{n 3^n}{n^3} = \frac{3^n}{n^2}$.
Does the benchmark series converge? We can use the Divergence Test on $b_n$: $\lim_{n \to \infty} \frac{3^n}{n^2} = \infty$. Since the terms do not go to zero, $\sum b_n$ diverges.
Now, apply the LCT: $\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \left( \frac{n 3^n}{4n^3+2} \cdot \frac{n^2}{3^n} \right) = \lim_{n \to \infty} \frac{n^3}{4n^3+2} = \frac{1}{4}$
Since $\frac{1}{4} > 0$ and finite, our original series shares the fate of the benchmark series, meaning it diverges.
Let an = $\frac{\sqrt{n}}{n-1}$
$\frac{\sqrt{n}}{n-1}>\frac{\sqrt{n}}{n}=\frac{1}{\sqrt{n}}$.
Let bn = $\frac{1}{\sqrt{n}}, \sum_{n=2}^\infty \frac{1}{\sqrt{n}}$ diverges by the Theorem (p-series) ⇒ $\sum_{n=2}^\infty \frac{\sqrt{n}}{n-1}$ also diverges by the Direct Comparison test.
As $n \to \infty$, the numerator $e^{1/n} \to e^0 = 1$. Therefore, the series behaves essentially like $\frac{1}{n}$. Let $b_n = \frac{1}{n}$ (the divergent harmonic series).
$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \left( \frac{e^{1/n}}{n} \cdot \frac{n}{1} \right) = \lim_{n \to \infty} e^{1/n} = e^0 = 1$
Since $1 > 0$ and finite, the original series $\sum_{n=1}^\infty \frac{e^{\frac{1}{n}}}{n}$ diverges by the LCT.
We compare this to the geometric series $\sum_{n=1}^\infty \frac{1}{3^n}$.
The Logic: For all $n \geq 1$, it is strictly true that $3^n + n^2 > 3^n$. When we take the reciprocal of both sides of a positive inequality, the direction of the inequality flips: $\frac{1}{3^n+n^2} < \frac{1}{3^n}$
The series $\sum_{n=1}^\infty \frac{1}{3^n}$ is a geometric series with a common ratio of $r = \frac{1}{3}$. Because $\vert{}r\vert{} < 1$, the geometric series converges. Since our original series consists of positive terms that are strictly smaller than the terms of a known convergent series, the Direct Comparison Test guarantees that $\sum_{n=1}^\infty \frac{1}{3^n+n^2}$ also converges.
For very large values of $n$, the $+n$ inside the square root becomes mathematically negligible compared to the $n^3$. If we strip away the non-dominant terms, the series behaves roughly like $\frac{n}{\sqrt{n^3}} = \frac{n}{n^{3/2}} = \frac{1}{n^{1/2}}$.
Comparison: We compare our original series $a_n$ to $b_n = \frac{1}{n^{1/2}}$. The series $\sum_{n=1}^\infty \frac{1}{n^{1/2}}$ diverges by the $p$-series test (since $p = \frac{1}{2} \leq 1$).
The Limit Computation: $\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \left( \frac{n}{\sqrt{n^3+n}} \cdot \frac{n^{1/2}}{1} \right) = \lim_{n \to \infty} \frac{n^{3/2}}{\sqrt{n^3+n}}$
To evaluate this, divide both the numerator and the denominator by $n^{3/2}$. Inside the square root, dividing by $n^{3/2}$ is equivalent to dividing by its square, $n^3$: $\lim_{n \to \infty} \frac{1}{\sqrt{\frac{n^3+n}{n^3}}} = \lim_{n \to \infty} \frac{1}{\sqrt{1 + \frac{1}{n^2}}} = \frac{1}{\sqrt{1 + 0}} = 1$
Conclusion: The calculated limit is $L = 1$. Since $0 < L < \infty$, the Limit Comparison Test tells us that both series share the exact same fate. Because our comparison series diverges, the original series diverges as well.
Suppose we have identical blocks of uniform density and length 2. The blocks must be supported entirely by their own weight: no glue, no external supports, and only one block at each level.
The optimal strategy is to build the stack from the top down, placing each new block as far out as possible while keeping the center of mass of everything above it directly over the supporting block.
Let $C_n$ denote the position of the center of mass of the top (n) blocks, measured from the left end of the top block.
This is the (N)-th harmonic number, $H_N$.
Using the integral estimate for the harmonic series, $\ln N < H_N < 1+\ln N$ and because $\lim_{N\to\infty}\ln N=\infty$, we obtain $\boxed{\lim_{N\to\infty}H_N=\infty}.$
Therefore, $\boxed{\text{there is no finite theoretical upper limit to the overhang.}}$ With sufficiently many blocks, the stack can extend arbitrarily far beyond the edge of the table. This is correct only in the ideal mathematical model. In reality, imperfections, friction, vibrations, deformation of the blocks, air currents, and measurement errors impose practical limits.
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