Sequences & Series of Real Numbers

Picture a pizza. You eat half, then half of what is left, then half of that, and you never stop. Each bite is tinier than the last, yet how much pizza do you eat altogether? Almost exactly one whole pizza. Adding up an endless list of shrinking numbers and still getting a sensible total is the whole idea of an infinite series. Sometimes the running total settles on a number (we say it converges); sometimes it just keeps growing (it diverges).

Call each number in the list a term $a_n$, so the list is $a_1, a_2, a_3, \dots$ The partial sum $S_N = a_1 + a_2 + \cdots + a_N$ is simply the running total after $N$ terms. The pizza is the geometric series $\sum (1/2)^n = \tfrac12 + \tfrac14 + \tfrac18 + \cdots$, whose partial sums step $0.5,\ 0.75,\ 0.875,\ \dots$ ever closer to $1$. Now the twist that surprises everyone: the harmonic series $\sum 1/n = 1 + \tfrac12 + \tfrac13 + \cdots$ has terms that also shrink to zero, yet its total grows without any ceiling. Shrinking terms alone are not enough to guarantee a finite answer.

Formally, $\sum a_n$ converges exactly when its partial sums $S_N$ approach a limit $L$ as $N \to \infty$. Two everyday tools settle most cases. The ratio test watches $L = \lim_{n\to\infty} \left| a_{n+1}/a_n \right|$: if $L < 1$ the terms fall off geometrically and the series converges; if $L > 1$ it diverges. The $p$-series rule states that $\sum 1/n^p$ converges if and only if $p > 1$ — so $\sum 1/n$ (here $p = 1$) diverges, while $\sum 1/n^2$ ($p = 2$) converges to the famous $\pi^2/6 \approx 1.645$. In the simulation, the Terms n slider sets how many terms $N$ you add, and the p slider sets the exponent $p$ used by the $p$-series preset.

Choose the Harmonic preset and drag Terms n all the way to 300 — watch $S_n$ keep creeping upward with no ceiling, divergence in slow motion. Then switch to the p-series preset with $p = 2$ and see the curve flatten against the dashed line at $\pi^2/6$. Finally drag $p$ below $1$: the limit line disappears and the sum runs away again, exposing the knife-edge right at $p = 1$.

$\sum1/n$ diverges despite $1/n\to0$. Proof by Cauchy condensation or grouping: $1+1/2+(1/3+1/4)+\cdots\geq1+1/2+1/2+\cdots$. This is the classic counterexample to "if $a_n\to0$ then $\sum a_n$ converges."

If $|a_{n+1}/a_n|\to L<1$, eventually $|a_{n+1}|1$: $|a_n|\to\infty$, diverges.

$\sum1/n^p$: converges for $p>1$ (integral test: $\int_1^\infty x^{-p}\,dx=1/(p-1)$). Diverges for $p\leq1$. Special cases: $p=1$ (harmonic, diverges), $p=2$ (Basel problem, $=\pi^2/6$), $p=3/2$ (converges).

Euler proved $\sum_{n=1}^\infty 1/n^2=\pi^2/6\approx1.6449$. Proof via Fourier series of $f(x)=x^2$ or via Parseval identity. The simulation shows $S_n\to\pi^2/6$ as $n\to\infty$: set $p=2$ in the preset.

Absolute: $\sum|a_n|$ converges (stronger). Conditional: $\sum a_n$ converges but $\sum|a_n|$ diverges. Alternating harmonic $\sum(-1)^{n+1}/n=\ln2$ is conditionally convergent. Remarkable: conditionally convergent series can be rearranged to any sum (Riemann rearrangement theorem).

$|a_n|=1/n^2$. $\sum1/n^2=\pi^2/6$ (converges, $p>1$). So the series converges absolutely. $\sum_{n=1}^\infty(-1)^{n+1}/n^2=\pi^2/12$ (from Fourier series). Partial sum with $n=50$: $S_{50}\approx0.8224$, exact $\pi^2/12\approx0.8225$. Error $<0.0001$.