show that $\sum_{i=1}^{n} (-1)^{i+1}{{n}\choose{i}}\frac{1}{i(i+1)} = \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n+1}$

Question

I have to show that

a) $\sum_{i=1}^{n} (-1)^{i+1}{{n}\choose{i}}\frac{1}{i(i+1)} = \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n+1}$

edit:

I no longer have to show that

$\sum_{i=0}^{n} (-1)^{i}{{n}\choose{i}}\frac{1}{i^2} = \frac{1}{n+1}(1+\frac{1}{2} + ... + \frac{1}{n})$

(as a second problem) because i managed to do it myself

Answer 1

We shall start with a couple of slightly easier sums \begin{eqnarray*} S_n=\sum_{i=1}^{n} (-1)^{i+1} \binom{n}{i} \frac{1}{i} \\ S_n^{+}=\sum_{i=1}^{n} (-1)^{i+1} \binom{n}{i} \frac{1}{i+1} \end{eqnarray*} Now use \begin{eqnarray*} \frac{1}{i}= \int_0^1 x^{i-1} dx \\ S_n=\sum_{i=1}^{n} (-1)^{i+1} \binom{n}{i} \int_0^1 x^{i-1} dx \end{eqnarray*} Invert the sum & integral. Use the binomial expansion ... \begin{eqnarray*} S_n=\int_0^1 \sum_{i=1}^{n} (-1)^{i+1} \binom{n}{i} x^{i-1} dx =\int_0^1 \frac{1-(1-x)^n}{x} dx \end{eqnarray*} Substitute $u=1-x$ & expand geometrically \begin{eqnarray*} S_n=\int_0^1 \frac{1-u^n}{1-x} du =\int_0^1 (1+u+ \cdots +u^{n-1})du =1+\frac{1}{2}+ \cdots +\frac{1}{n} \end{eqnarray*} Now do the second sum in a similar manner. \begin{eqnarray*} S_n^{+}=\int_0^1 (1-(1-x)^{n})dx=1-\frac{1}{n+1}. \end{eqnarray*} Partial fractions gives \begin{eqnarray*} \frac{1}{i(i+1)}=\frac{1}{i}-\frac{1}{i+1}. \end{eqnarray*} Your original sum is $S_n-S_n^+$ & the result follows.