>[!abstract] >Triangular numbers are the sequence of natural numbers $\{1, 3, 6, 10, 15, \cdots\}$ representing the total count of dots needed to form an equilateral triangle. Starting at the top of the triangle with one dot, each subsequent row below it adds one additional dot. The value of any triangular number on row $n$ can be found using the formula: > >$T_n = \frac{n(n+1)}{2}$ ## Visualization ``` • 1 • • + 2 = 3 • • • + 3 = 6 • • • • + 4 = 10 • • • • • + 5 = 15 • • • • • • + 6 = 21 • • • • • • • + 7 = 28 ``` ## Problem In 1672, Dutch mathematician and astronomer Christiaan Huygens challenged his student, German [[polymath]] Gottfried Wilhelm Leibniz, to find the exact sum of the infinite series of the reciprocals of triangular numbers, expressed as: $S_n = \sum_{n=1}^{\infty} \frac{1}{T_n} = 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \cdots$ ## Proof Leibniz found that $S_n = 2$ as $n \to \infty$ using the following proof. ### Step 1: Start with the reciprocal Start with the reciprocal of the formula for triangular numbers shown in the abstract: $\frac{1}{T_n} = \frac{2}{n(n+1)}$ ### Step 2: Split the reciprocal fraction Then split that fraction using partial fraction decomposition; to do so, we want to find $A$ and $B$ such that: $\frac{1}{T_n} = \frac{2}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}$ Putting $A$ and $B$ together using a common denominator: $\frac{A}{n} + \frac{B}{n+1} = \frac{A(n+1) + Bn}{n(n+1)}$ We can rewrite the numerator as follows: $\frac{A(n+1) + Bn}{n(n+1)} = \frac{An + A + Bn}{n(n+1)} = \frac{(A + B)n + A}{n(n+1)}$ So in order for the following equivalence to be true: $\frac{2}{n(n+1)} = \frac{(A + B)n + A}{n(n+1)}$ It must follow that: $(A + B)n + A = 2$ Since the right side has no $n$ terms: $(A + B)n = 0$ which means that: $A = 2$ and $B = -2$ Now, substituting back: $\frac{A}{n} + \frac{B}{n+1} = \frac{2}{n} - \frac{2}{n+1}$ Which we can finally factor into: $2 (\frac{1}{n} - \frac{1}{n+1})$ So we have now rewritten the reciprocal of a triangular number from step 1: $\frac{1}{T_n} = \frac{2}{n(n+1)}$ as $\frac{1}{T_n} = 2 (\frac{1}{n} - \frac{1}{n+1})$ ### Step 3: Apply to the series So now the entire infinite series of the reciprocals of triangular numbers $S_n$ becomes: $S_n = \sum_{n=1}^{\infty} \frac{1}{T_n} = 2 \sum_{n=1}^{\infty} (\frac{1}{n} - \frac{1}{n+1})$ Let's expand the first few terms: $2[(1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + \cdots]$ We can see that all the $\frac{1}{n}$ terms cancel each other out, leaving only: $S_n = 2 \sum_{n=1}^{\infty} (\frac{1}{n} - \frac{1}{n+1}) = 2(1) = 2$ So as $n \to \infty$, we conclude that: $S_n = \sum_{n=1}^{\infty} \frac{1}{T_n} = 2$ ## Visualization of the convergence ```R n <- 1:100 s <- cumsum(2 / (n * (n + 1))) par(bg = NA) plot(n, s, bg = NA, type = "l", lwd = 2, xlab = "Number of terms", ylab = "Partial sum", main = "Convergence of Σ(1/Tn) to 2") abline(h = 2, lty = 2, lwd = 2) ``` ![[Triangular numbers graph.png]] >[!related] >- **North** (upstream): — >- **West** (similar): — >- **East** (different): — >- **South** (downstream): —