In this problem, we will prove that the ``average'' number of divisors
of a number of size about $n$ is ``about'' $\ln n$.  For formally, let
$t(n)$ be the number of divisors of the number $n$, and let
\[
T(n) = \frac1n \sum_{j=1}^n t(j)
\]
be the average number of divisors of all numbers less than or equal to
$n$.  Show that $T(n) \sim \ln n$.  {\bf Hint:} let the function
$f(i,j)=1$ if $i$ is a divisor of $j$, and $0$ otherwise.  What do the
row and column sums of $f(i,j)$ mean?


What is wrong with the following summation?
\[
(\sum_{j=1}^n a_j)(\sum_{k=1}^n (1/a_k) =
\sum_{j=1}^n \sum_{k=1}^n a_j/a_k =
\sum_{k=1}^n \sum_{k=1}^n a_k/a_k =
\sum_{k=1}^n \sum_{k=1}^n 1 =
\sum_{k=1}^n n = n^2
\]

Riemann's Zeta Function $\zeta(k)$ is defined to be the infinite sum
\[
1+\frac1{2^k}+\frac1{3^k} \cdots = \sum_{j \ge 1} \frac1{j^k}
\]
Prove that 
\[
\sum_{k \ge 2} (\zeta(k)-1) = 1
\]
{\bf Hint: You may use (without proof) problem 2 from set 5.}


% Local Variables: 
% mode: latex
% TeX-master: t
% End: 
