Question

h_1 = 0

h_{n+1} = (n+1) * h_{n} + n!

Find an explicit formula for a generating function of h_n.

Use the formula to prove that h_{n} = n! * SUM{from k =1 to n} 1/k.

Answer 1

We want to prove that the solution of $(A)\quad h_{n+1} =(n+1)\times h_n + n!\,\text{with} \,h_1= 0$ is $h_n = n!\times\,\sum _{k=1}^{n}\frac 1k$

But note that the required solution does not satisfy the given initial condition . The initial condition should be .

Here we first write down two or three successive terms, then guess a pattern. Finally we invoke the principle of mathematical induction to assert that our guess is the correct solution of the given problem.

So here initial condition is , put n=1 in (A)

$h_2 = 2\cdot h_1 + 1! = 2\cdot 1 + 1! =2! + 2!/2 = 2!(1+\frac12) =2!\sum _{k=1}^2\frac1k$

put n=2 in (A)

$h_3 = 3\cdot h_2 + 2! = 3\cdot (2\cdot 1+ 1!) + 2! =3! + 3\times 1! + 2! = 3!(1+\frac12+ \frac 13) =3!\sum _{k=1}^3\frac1k$

So this pattern gives a possible solution to the problem. Assume that the pattern is true for n, That is,

$h_n = n!\times\,\sum _{k=1}^{n}\frac 1k$

Now we shall prove that the pattern is true for n+1

Consider $h_{n+1} = (n+1)\cdot h_n + n! = (n+1)\times[n!\times \sum _{k=1}^n\frac1k] +n! = (n+1)\times n!)\sum _{k=1}^n\frac1k + n!$

$h_{n+1} = (n+1)!\bigg [\sum _{k=1}^n\frac1k+\frac{1}{n+1}\bigg] = (n+1)!\bigg [\sum _{k=1}^{n+1}\frac1k\bigg]$

So by the principle of mathematical induction,the result is true for all integral values of n.