The limit of $\lim_{n \to +\infty} a_0+a_1+a_2+.......a_n$ where $a_p=\sum_{i=0}^p (-1)^i\frac{\binom{p}{i}}{(i+2)(i+4)}$

Question

I have taken this question from molodovian national MO 2008 The question is as follows

The sequence $(a_p)_p\ge 0$ is defined as $$a_p=\sum_{i=0}^p (-1)^i\frac{\binom{p}{i}}{(i+2)(i+4)}$$

Now let's find the limit

\begin{align*} \lim_{n \to +\infty} a_0+a_1+a_2+.......a_n \end{align*}

Can anyone help me to go further.

Answer 1

Let $$\begin{split} S_{k,p} &= \sum_{i=0}^p\binom{p}{i}\frac{(-1)^i}{i+k}\\ &= \sum_{i=0}^p\binom{p}{i}(-1)^i\int_0^1 x^{i+k-1}dx\\ &= \int_0^1 \sum_{i=0}^p\binom{p}{i}(-1)^ix^{i+k-1}dx\\ &= \int_0^1 x^{k-1}(1-x)^pdx \end{split}$$ Then since $$\frac 1{(i+2)(i+4)}=\frac 1 2 \left(\frac 1 {i+2}-\frac 1 {i+4}\right)$$ we have $$a_p=\frac 1 2 (S_{2,p}-S_{4,p})=\frac 1 2 \int_0^1 (x-x^3)(1-x)^pdx$$ Thus,

$$\begin{split} \sum_{p=0}^n &a_p= \frac 1 2 \int_0^1(x-x^3)\sum_{p=0}^n(1-x)^pdx\\ &=\frac 1 2 \int_0^1(x-x^3)\frac{1-(1-x)^{n+1}}{x}dx\\ &=\frac 1 2 \int_0^1 (1-x^2)(1-(1-x)^{n+1})dx\\ \end{split}$$

Using Lebesgue's Dominated Convergence Theorem: $$\lim_{n\rightarrow+\infty}\sum_{p=0}^n a_p = \frac 1 2 \int_0^1 (1-x^2)dx=\frac 1 2 \left (1- \frac 1 3\right) = \frac 1 {3}$$

Answer 2

The solution of Stefan Lafon is elegant and shows the usefulness of professional tools. Integral representations (or, moreover, the dominated convergence theorem) are probably not the intended solution of an MO problem, though. So it may be interesting to see how to derive explicit expressions for $a_p$ and the partial sums without any calculus.

This can be done with finite differences (https://en.wikipedia.org/wiki/Finite_difference): If we define $\Delta_h[f](x)=f(x+h)-f(x)$, the formula for $n$th order forward differences will be $$\Delta^n_h[f](x)=\sum^n_{i=0}\,(-1)^{n-i}\binom{n}{i}f(x+i\,h),$$ or for the special case of stepsize $h=1$, $$\Delta^n[f](x)=\sum^n_{i=0}\,(-1)^{n-i}\binom{n}{i}f(x+i).$$ This means that $$a_p=(-1)^p\Delta^n[f](0),\tag1$$ where $\displaystyle f(x)=\frac1{(x+2)(x+4)}$. Now it's easy to calculate forward differences for $\displaystyle f_a(x)=\frac1{x+a}$: $$\Delta[f_a](x)=\frac1{x+a+1}-\frac1{x+a}=\frac{-1}{(x+a)(x+a+1)},$$ $$\Delta^2[f_a](x)=\frac{-1}{(x+a+1)(x+a+2)}-\frac{-1}{(x+a)(x+a+1)}=\frac{(-1)(-2)}{(x+a)(x+a+1)(x+a+2)},$$ and by induction $$\Delta^p[f_a](x)=\frac{(-1)^p\,p!}{(x+a)\ldots(x+a+p)}.$$ For a positive integer $a$, this gives $$\Delta^p[f_a](0)=\frac{(-1)^p\,(a-1)!\,p!}{(p+a)!}=(-1)^p\,\frac{(a-1)!}{(p+1)\ldots(p+a)}.\tag2$$ Since $$f(x)=\frac1{(x+2)(x+4)}=\frac12\,\left(f_2(x)-f_4(x)\right),$$ combining (1) and (2) yields $$a_p=\frac12\,\left(\frac{1!}{(p+1)(p+2)}-\frac{3!}{(p+1)(p+2)(p+3)(p+4)}\right)=\frac{p+6}{2\,(p+2)(p+3)(p+4)},\tag3$$ as announced in a comment. Of course, it would be possible to obtain (3) from the integral representation as well, substituting $x\to 1-y$. The partial sums $\displaystyle\sum^n_{p=0}a_p$ can be computed as telescoping sums, using $$\frac1{(p+1)(p+2)}=\frac1{p+1}-\frac1{p+2}$$ and $$\frac1{(p+1)(p+2)(p+3)(p+4)}=\frac13\left(\frac1{(p+1)(p+2)(p+3)}-\frac1{(p+2)(p+3)(p+4)}\right).$$ So, $$\sum^n_{p=0}a_p=\frac12\left(1-\frac1{n+2}\right)-\frac16+\frac1{(n+2)(n+3)(n+4)}=\frac{2n^2+11n+9}{6\,(n+3)(n+4)}.$$