341Hey, I couldnt think of an induction proof :D.
This is fun to work out. Let \tau(m) represent the number of divisors of the natural number m.
Then prove that
\tau(1)+\tau(1)+ ...+\tau(n) = \left[\frac{n}{1} \right] + \left[\frac{n}{2} \right]+...+\left[\frac{n}{n} \right]
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3 Answers
62Hint: Induction...
(i couldnt think of a direct algebraic method so far :( )
62we simply have to prove that \tau (n+1) =\left\{\left[\frac{n+1}{1} \right] + \left[\frac{n+1}{2} \right]+...+\left[\frac{n+1}{n+1} \right] \right\} - \left\{\left[\frac{n}{1} \right] + \left[\frac{n}{2} \right]+...+\left[\frac{n}{n} \right] \right\}
which i think is directly obvious is we club the whole summation as
\tau (n+1) =\left\{\left[\frac{n+1}{1} \right] - \left[\frac{n}{1}\right]\right\}+\left\{\left[\frac{n+1}{2} \right] - \left[\frac{n}{2}\right]\right\}+...+\left\{\left[\frac{n+1}{n+1} \right] - \left[\frac{n}{n}\right]\right\}
The logic is that the number of factors of n+1 will be contributed by the fact that N+1 is a multiple of r or not ...