## Saturday, September 21, 2013

### $\sum_{k=1}^{n} H_k = (n+1)(H_{n+1} - 1)$

$$\frac{1}{1-z} = 1 + z + z^2 + z^3 \dots \\ \int \frac{1}{1-z}dz = \frac{z}{1} + \frac{z^2}{2} + \frac{z^3}{3} \dots = \sum_{k=1}^{\infty} \frac{z}{k} z^k = -\ln(1-z)\\ \\ H_n = \sum_{k=1}^{n} \frac{1}{k} \hspace{0.5in} \mbox{By definition, also notice they are partial sums}\\ \rightarrow \mbox{O.G.F of H_n is}\hspace{0.02in} \frac{1}{1-z}\times (-\ln(1-z)) = \frac{1}{1-z}\ln(\frac{1}{1-z}) \hspace{0.5in}\mbox{from our previous result on partial sums}\\ \rightarrow \frac{1}{1-z}\ln(\frac{1}{1-z}) = H_1\times z + H_2\times z^2 + H_3\times z^3 \ldots \hspace{0.5in} \mbox{(1)}\\ \rightarrow \frac{d}{dz}(\frac{1}{1-z}\ln(\frac{1}{1-z})) = \frac{1}{(1-z)^2}\ln(\frac{1}{1-z}) + \frac{1}{(1-z)^2} = H_1 + 2H_2\times z + 3H_3\times z^2 \ldots \hspace{0.5in}\mbox{(2)}\\ \mbox{Notice that}\hspace{0.02in} \frac{1}{(1-z)^2} = 1 + 2z + 3z^2 \dots \hspace{0.5in} \mbox{(3)}\\ \rightarrow\mbox{(2) - (3)} = \frac{1}{(1-z)^2}\ln(\frac{1}{(1-z)}) = (H_1-1) + (2H_2-2)\times z + (3H_3 -3) \times z^2 \ldots \\ \rightarrow \frac{1}{(1-z)^2}\ln(\frac{1}{(1-z)}) \hspace{0.02in} \mbox{is the O.G.F of sequence} \hspace{0.02in} (n+1)H_{n+1} - (n+1) = (n+1)(H_{n+1}-1) \hspace{0.5in} \mbox{(4)}\\ \\ \mbox{Applying the result of the partial sums to (1) the O.G.F of sequence}\hspace{0.02in} \sum_{k=1}^{n} H_n \hspace{0.02in} \mbox{is} \hspace{0.02in} \frac{1}{(1-z)}\times {\frac{1}{(1-z)}\ln(\frac{1}{1-z})} = \frac{1}{(1-z)^2}\ln(\frac{1}{(1-z)}) \hspace{0.5in} \mbox{(5)}\\ \mbox{From (4) and (5) the generating functions of both the sequences are same, hence the corresponding terms must be the same}$$

### Simple proof for the O.G.F of partial sums

Let $A(z)$ be the O.G.F of the sequence $S = \{a_0, a_1, a_2 \ldots \}$, then we can derive the O.G.F of the sequence $b_n = \sum_{i=0}^{n} a_i$ (i.e. $B = \{a_0, a_0+a_1, a_0+a_1+a_2 \ldots \}$) as follows.

Consider the sequence $S^1 = \{0 , a_0, a_1, a_2, a_3 \ldots \}$ then O.G.F corresponding to $S^1$ is $0\times z^0 + a_0 \times z + a_1 \times z^2 + a_2 \times z^3 \ldots = z\times A(z)$. Similarly the O.G.F corresponding to the sequence $S^2 = \{0, 0, a_0, a_1, a_3 \ldots \}$ is $z^2 \times A(z)$. Now notice that if we add corresponding terms in the sequences $S^1 , S^2, S^3 \ldots$ then this
results in the sequence $B$. Hence the generating function corresponding to $B$ is $A(z) + z\times A(z) + z^2\times A(z) + z^3\times A(z) \ldots = A(z)\times(1 + z + z^2 + z^3 \ldots) = \frac{A(z)}{1-z}$ $\Box$.