More generally, an integer $N$ is divisible by $3$ if and only if the sum of the digits appearing in $N$ is divisible by $3$.

## Tuesday, November 12, 2013

### Simple Divisibility Tests

In this post we will justify some common divisibility tests that most
school children are familiar with. Recall that a number such as
$111$ is divisible by $3$ if and only if the sum of the digits is
divisible by $3$. Since $1+1+1=3$ is divisible by $3$,
we see that $111$ is divisible by three.

More generally, an integer $N$ is divisible by $3$ if and only if the sum of the digits appearing in $N$ is divisible by $3$.

More generally, an integer $N$ is divisible by $3$ if and only if the sum of the digits appearing in $N$ is divisible by $3$.

## Saturday, November 2, 2013

### A Simple Proof that the Harmonic Series Diverges

One usually encounters the harmonic series
\[
\sum_{k=1}^{\infty} \frac{1}{k}
\]
as an example of a series that diverges for non-obvious reasons.
With $H_n$ defined to be the partial sum $H_n:=\sum_{k=1}^n \frac{1}{k}$,
a typical way to prove divergence is to observe that
\begin{align*}
H_{2n} &= H_n + \frac{1}{n+1} + \dots + \frac{1}{2n} \\
& \geq H_n + \underbrace{\frac{1}{2n} + \dots + \frac{1}{2n}}_{n \text{ terms}} \\
& = H_n + \frac{1}{2}.
\end{align*}
In particular, $H_{2^n} \geq 1 + \frac{n}{2}$, so the sequence of partial sums of
the harmonic series has an unbounded subsequence, whence the series diverges.

Subscribe to:
Posts (Atom)