$$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$

$$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$

Inline double escaped: \(x^2+1\)

Block dollars:

$$
\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}
$$

One of the first beautiful surprises in analysis is that the harmonic series diverges, even though its terms go to zero.

The series is

$$
1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots
$$

At first it feels like it should converge, because the terms (1/n) become smaller and smaller. But the condition (a_n \to 0) is necessary for convergence, not sufficient.

The classic grouping argument is very neat:

$$
1 + \frac{1}{2}

• \left(\frac{1}{3} + \frac{1}{4}\right)
• \left(\frac{1}{5} + \cdots + \frac{1}{8}\right)
• \left(\frac{1}{9} + \cdots + \frac{1}{16}\right)
• \cdots
  $$

Now look at each group.

For the group

$$
\frac{1}{3} + \frac{1}{4},
$$

both terms are at least (1/4), so

$$
\frac{1}{3} + \frac{1}{4} \geq \frac{1}{4} + \frac{1}{4} = \frac{1}{2}.
$$

For the next group,

$$
\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8},
$$

all four terms are at least (1/8), so

$$
\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}
\geq 4 \cdot \frac{1}{8}
= \frac{1}{2}.
$$

In general, the group

$$
\frac{1}{2^k+1} + \frac{1}{2^k+2} + \cdots + \frac{1}{2^{k+1}}
$$

contains (2^k) terms, and each term is at least (1/2^{k+1}). Therefore the whole group is at least

$$
2^k \cdot \frac{1}{2^{k+1}} = \frac{1}{2}.
$$

So the harmonic series is bounded below by

$$
1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \cdots
$$

which clearly diverges.

Therefore,

$$
\sum_{n=1}^{\infty} \frac{1}{n} = \infty.
$$

The moral is: terms going to zero is not enough. The terms must go to zero fast enough.

One of the first beautiful surprises in analysis is that the harmonic series diverges, even though its terms go to zero.

The series is

$$
1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots
$$

At first it feels like it should converge, because the terms (1/n) become smaller and smaller. But the condition (a_n \to 0) is necessary for convergence, not sufficient.

The classic grouping argument is very neat:

$$
1 + \frac{1}{2}

• \left(\frac{1}{3} + \frac{1}{4}\right)
• \left(\frac{1}{5} + \cdots + \frac{1}{8}\right)
• \left(\frac{1}{9} + \cdots + \frac{1}{16}\right)
• \cdots
  $$

Now look at each group.

For the group

$$
\frac{1}{3} + \frac{1}{4},
$$

both terms are at least (1/4), so

$$
\frac{1}{3} + \frac{1}{4} \geq \frac{1}{4} + \frac{1}{4} = \frac{1}{2}.
$$

For the next group,

$$
\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8},
$$

all four terms are at least (1/8), so

$$
\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}
\geq 4 \cdot \frac{1}{8}
= \frac{1}{2}.
$$

In general, the group

$$
\frac{1}{2^k+1} + \frac{1}{2^k+2} + \cdots + \frac{1}{2^{k+1}}
$$

contains (2^k) terms, and each term is at least (1/2^{k+1}). Therefore the whole group is at least

$$
2^k \cdot \frac{1}{2^{k+1}} = \frac{1}{2}.
$$

So the harmonic series is bounded below by

$$
1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \cdots
$$

which clearly diverges.

Therefore,

$$
\sum_{n=1}^{\infty} \frac{1}{n} = \infty.
$$

The moral is: terms going to zero is not enough. The terms must go to zero fast enough.