Proving Binomial Identity without calculus

Question

How to establish the following identities without the help of calculus:

For positive integer $n, $ $$\sum_{1\le r\le n}\frac{(-1)^{r-1}\binom nr}r=\sum_{1\le r\le n}\frac1r $$

and $$\sum_{0\le r\le n}\frac{(-1)^r\binom nr}{4r+1}=\frac{4^n\cdot n!}{1\cdot5\cdot9\cdots(4n+1)}$$

Answer 1

For the first Question let $\displaystyle S_n=\sum_{1\le r\le n}\frac{(-1)^{r-1}\binom nr}r$

$\displaystyle\implies S_{m+1}-S_m=\sum_{1\le r\le m+1}(-1)^{r-1}\frac{\binom{m+1}r-\binom mr}r-\binom m{m+1}\frac{(-1)^m}{m+1}$

$\displaystyle=\sum_{1\le r\le m+1}(-1)^{r-1}\frac{\binom{m+1}r-\binom mr}r$ as $\binom mr=0$ for $r>m$ or $r<0$

Now using this formula, $\displaystyle\binom{m+1}r=\binom mr+\binom m{r-1}\iff \binom{m+1}r-\binom mr=\binom m{r-1}$

Again, $\displaystyle\frac{\binom m{r-1}}r=\frac{m!}{\{m-(r-1)\}!(r-1)!\cdot r}=\frac1{m+1}\cdot\frac{(m+1)!}{(m+1-r)!\cdot r!}$ $\displaystyle=\frac1{m+1}\cdot\binom{m+1}r$

$\displaystyle\implies S_{m+1}-S_m=\sum_{1\le r\le m+1}(-1)^{r-1}\cdot\frac1{m+1}\cdot\binom{m+1}r$ $\displaystyle=\frac1{m+1}\sum_{1\le r\le m+1}(-1)^{r-1}\binom{m+1}r$ $\displaystyle=\frac{1-(1-1)^{m+1}}{m+1}$

$\displaystyle\implies S_{m+1}-S_m=\frac1{m+1}$

Now, $\displaystyle S_1=\frac11$

As $\displaystyle S_2-S_1=\frac12\implies S_2=1+\frac12$ and so on

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\LARGE\left. {\bf 1}\right):} \bbox[15px,#ffd]{\ds{\sum_{1\ \leq\ r\ \leq\ n}{\pars{-1}^{\,r - 1}{n \choose r} \over r} = \sum_{1\ \leq\ r\ \leq\ n}{1 \over r}:\ {\large ?}}}$. \begin{align} \sum_{1\ \leq\ r\ \leq\ n}{\pars{-1}^{\,r-1}{n \choose r} \over r} & = \sum_{r = 1}^{n}\pars{-1}^{\,r - 1}{n \choose r}\int_{0}^{1}t^{r - 1}\,\dd t = -\int_{0}^{1}\sum_{r = 1}^{n}{n \choose r}\pars{-t}^{\,r}\,{\dd t \over t} \\[5mm] & = -\int_{0}^{1}{\pars{1 - t}^{n} - 1 \over t}\,\dd t = \int_{0}^{1}{t^{n} - 1 \over t - 1}\,\dd t = \int_{0}^{1}\sum_{r = 1}^{n}t^{r - 1}\,\dd t \\[5mm] & = \sum_{r = 1}^{n}\int_{0}^{1}t^{r - 1}\,\dd t = \bbx{\sum_{1\ \leq\ r\ \leq\ n}{1 \over r}} \\ & \end{align}

---

$\ds{\LARGE\left. {\bf 2}\right):} \bbox[15px,#ffd]{\ds{% \sum_{0\ \leq\ r\ \leq\ n}{\pars{-1}^{r}{n \choose r} \over 4r + 1} = {4^{n}\,n! \over 1 \cdot 5 \cdot 9 \cdots \pars{4n + 1}}:\ {\large ?}}}$. \begin{align} \sum_{0\ \leq\ r\ \leq\ n}{\pars{-1}^{\,r}\,{n \choose r} \over 4r + 1} & = \sum_{r = 0}^{n}\pars{-1}^{\,r}{n \choose r}\int_{0}^{1}t^{4r} \,\dd t = \int_{0}^{1}\sum_{r = 0}^{n}{n \choose r}\pars{-t^{4}}^{r}\,\dd t \\[5mm] & = \int_{0}^{1}\pars{1 - t^{4}}^{n}\,\dd t \,\,\,\stackrel{\large t^{4}\ \mapsto\ t}{=}\,\,\, {1 \over 4}\int_{0}^{1}t^{-3/4}\,\pars{1 - t}^{n}\,\dd t \\[5mm] & = {1 \over 4}\,{\Gamma\pars{1/4}\Gamma\pars{n + 1} \over \Gamma\pars{n + 5/4}} = {1 \over 4}\,n!\,{1 \over \Gamma\pars{1/4 + \bracks{n + 1}}/\Gamma\pars{1/4}} \\[5mm] & = {1 \over 4}\,n!\,{1 \over \pars{1/4}^{\,\overline{n + 1}}} = {1 \over 4}\,n!\,{1 \over \prod_{r = 0}^{n}\pars{1/4 + r}} \\[5mm] & = {1 \over 4}\,n!\,{1 \over \prod_{r = 0}^{n}\bracks{\pars{4r + 1}/4}} \\[5mm] & = {1 \over 4}\,n!\,{1 \over \bracks{\prod_{r = 0}^{n}\pars{4r + 1}}/4^{n + 1}} = \bbx{\ds{{4^{n}\,n! \over 1 \cdot 5 \cdot 9 \cdots \pars{4n + 1}}}} \\ & \end{align}

Answer 3

Here is a purely combinatorial proof of the first identity:

Let us count the number $C$ of disjoint cycles in a given permutation $\sigma \in S_n$. On the one hand, we have simply $$C = \sum_{k=1}^{n} C_k$$ where $C_k$ is the number of $k$-cycles in $\sigma$. On the other hand, we can count by "inclusion-exclusion," as follows.

Denote the set $[n]:= \{1,2, 3, \cdots, n\}$. For each subset $S \subset [n]$, let $\chi(S) = 1$ if $S$ is contained wholly in some cycle of $\sigma$ and $0$ otherwise. Then, I claim we have $$C = \sum_{\emptyset \neq S \subset [n]} (-1)^{|S|-1} \chi(S)$$ To prove this, note that the only subsets of $[n]$ contributing to the sum are those contained in some cycle. For each cycle in $\sigma$ of size $k$, the contribution to the sum is $$\binom{k}{1} - \binom{k}{2} + \binom{k}{3} -\cdots + (-1)^{k-1} \binom{k}{k}= 1-(1-1)^k = 1$$ by the binomial theorem; hence, the sum ultimately evaluates to just the number of cycles of $\sigma$.

Thus, we have the identity $$\sum_{k=1}^{n} C_k = \sum_{\emptyset \neq S \subset [n]} (-1)^{|S|-1} \chi(S) = \sum_{k=1}^{n} (-1)^{k-1} \sum_{\substack{S \subset [n] \\ |S| = k}} \chi(S)$$ Now, we suppose $\sigma$ is chosen uniformly at random from $S_n$ and compute the expected number of cycles in $\sigma$. By linearity of expectation, we have $$\sum_{k=1}^{n} \mathbb{E}[C_k] = \sum_{k=1}^{n} (-1)^{k-1}\mathbb{E} \left[ \sum_{\substack{S \subset [n] \\ |S| = k}} \chi(S) \right]$$

Now, we have to compute these expectations. Let us first calculate $\mathbb{E}[C_k]$. We have $$C_k = \frac{1}{k} \sum_{j=1}^{n} \mathbb{1}_{\text{contained in a } k-\text{cycle}} (j) \implies \mathbb{E}[C_k] = \frac{n}{k} \text{Prob}(\text{fixed element contained in } k-\text{cycle})$$ and $$\mathbb{E} \left[ \sum_{\substack{S \subset [n] \\ |S| = k}} \chi(S) \right] = \binom{n}{k} \text{Prob}(\text{fixed set of size } k \text{ contained in some cycle})$$

The evaluation of these probabilities is classical (see Terry Tao's post or this Wikipedia article), and we have $$\text{Prob}(\text{fixed element contained in } k-\text{cycle}) = \frac{1}{n}$$ $$\text{Prob}(\text{fixed set of size } k \text{ contained in some cycle}) = \frac{1}{k}$$ hence the final conclusion follows.

Answer 4

You can always use Petkovsek's algorithm. It only requires some algebra to prove this and other problems alike.

You can read about it in the book $A=B$ (available free online).

Another thing is that derivation of polynomials is a completely algebraic operation.

You can always write instead of $(P(x))'|_{x=1}$ write $[P(x+1)-P(1)]/x|_{x=0}$, perhaps what is equivalent, rewrite in powers of $(x-1)$, which involves iterated division by $(x-1)$. [I just pressed Alt+F7 to try to compile the LaTeX] And little by little hide the calculus from the proof that you have.

Answer 5

Here is a probabilistic proof of the first equality.

Given $n$ urns, and each round, you place a ball into one of the $n$ urns randomly and uniformly. What is the expected number of rounds until every urn has a ball?

There are two approaches to solving this.

One approach uses inclusion-exclusion, which gives $$\sum_{k=1}^n(-1)^{k-1}\binom{n}{k}\frac{n}{k}.$$

The other approach let $X_k$ be the number of rounds to get balls in $k$ urns after you've gotten a ball in each of $k-1$ urns. Then it is easy to see that $E(X_k)=\frac{n}{n-k+1}$ and the expected time is $\sum_{k=1}^{n}\frac{n}{n-k+1}$ That can be rewritten as $$\sum_{k=1}^{n}\frac{n}{k}.$$

See this answer for details.

Answer 6

A different way to solve the first question following an algebraic approach might go as follows.

We have $$ S_{\,1} (n) = \sum\limits_{1\, \le \,\,r\, \le \,n} {\left( { - 1} \right)^{\,r - 1} \;{1 \over r}\left( \matrix{ n \cr r \cr} \right)} \quad \quad S_{\,2} (n) = \sum\limits_{1\, \le \,\,r\, \le \,n} {{1 \over r}} $$

Let's take the finite difference of $S_1$ $$ \eqalign{ & \Delta S_{\,1} (n) = S_{\,1} (n + 1) - S_{\,1} (n) = \cr & = \sum\limits_{1\, \le \,\,r\, \le \,n + 1} {\left( { - 1} \right)^{\,r - 1} \;{1 \over r}\left( \matrix{ n + 1 \cr r \cr} \right)} - \sum\limits_{1\, \le \,\,r\, \le \,n} {\left( { - 1} \right)^{\,r - 1} \;{1 \over r}\left( \matrix{ n \cr r \cr} \right)} = \quad\quad (0) \cr & = \sum\limits_{1\, \le \,\,r\, \le \,n + 1} {\left( { - 1} \right)^{\,r - 1} \; {1 \over r}\left( {\left( \matrix{ n + 1 \cr r \cr} \right) - \left( \matrix{ n \cr r \cr} \right)} \right)} = \quad \quad (1) \cr & = \sum\limits_{1\, \le \,\,r\, \le \,n + 1} {\left( { - 1} \right)^{\,r - 1} \;{1 \over r}\left( \matrix{ n \cr r - 1 \cr} \right)} = \quad \quad (2) \cr & = {1 \over {n + 1}}\sum\limits_{1\, \le \,\,r\, \le \,n + 1} {\left( { - 1} \right)^{\,r - 1} \;\left( \matrix{ n + 1 \cr r \cr} \right)} = \quad \quad (3) \cr & = {1 \over {n + 1}}\left( { - \left( { - 1} \right) - \sum\limits_{0\, \le \,\,r\, \le \,n + 1} {\left( { - 1} \right)^{\,r} \;\left( \matrix{ n + 1 \cr r \cr} \right)} } \right) = \quad \quad (4) \cr & = {1 \over {n + 1}}\quad \quad (5) \cr} $$ where:
- (0) we can extend the limit of the 2nd sum;
- (1) join the sums;
- (2) recursion of binomial;
- (3) absorption identity;
- (4) minus plus the term $r=0$;
- (5) the sum is null.

Since it is $$ \left\{ \matrix{ S_{\,1} (1) = S_{\,2} (1) = 1 \hfill \cr \Delta S_{\,1} (n) = \Delta S_{\,2} (n) = {1 \over {n + 1}} \hfill \cr} \right. $$ then the two sums are equal.

Concerning the second question, we rewrite the sum as an iterated Finite Difference $$ \eqalign{ & S_{\,3} (n) = \sum\limits_{\left( {0\, \le } \right)\,\,r\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,r} \;{1 \over {1 + 4r}}\left( \matrix{ n \cr r \cr} \right)} = \cr & = {{\left( { - 1} \right)^{\,n} } \over 4}\sum\limits_{\left( {0\, \le } \right)\,\,r\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,n - r} \;{1 \over {1/4 + r}}\left( \matrix{ n \cr r \cr} \right)} = \cr & = {{\left( { - 1} \right)^{\,n} } \over 4}\left. {\Delta ^{\,n} \left( {{1 \over {1/4 + x}}} \right)\,} \right|_{\,x\, = \,0} \cr} $$

Then we pass and consider the Falling and Rising Factorials for which the delta has a simple expression $$ \Delta _{\,x} ^m \;x^{\,\underline {\,n\,} } = n^{\,\underline {\,m\,} } x^{\,\underline {\,n - m\,} } $$ and for which applies the following indentity $$ x^{\underline {\, - m\,} } = {1 \over {\left( {x + m} \right)^{\underline {\,m\,} } }} = {1 \over {\left( {x + 1} \right)^{\overline {\,m\,} } }} $$ Therefore we can write $$ {1 \over {1/4 + x}} = {1 \over {\left( {1/4 + x} \right)^{\overline {\,1\,} } }} = \left( { - 3/4 + x} \right)^{\underline {\, - 1\,} } $$ and $$ \eqalign{ & \Delta ^{\,n} \left( {{1 \over {1/4 + x}}} \right) = \Delta ^{\,n} \left( { - 3/4 + x} \right)^{\underline {\, - 1\,} } = \left( { - 1} \right)^{\underline {\,n\,} } \left( { - 3/4 + x} \right)^{\underline {\, - 1 - n\,} } = \cr & = \left( { - 1} \right)^{\,n} n!\left( { - 3/4 + x} \right)^{\underline {\, - 1 - n\,} } \cr} $$ So the sum becomes $$ \eqalign{ & S_{\,3} (n) = {{\left( { - 1} \right)^{\,n} } \over 4}\left. {\Delta ^{\,n} \left( {{1 \over {1/4 + x}}} \right)\,} \right|_{\,x\, = \,0} = \cr & = {{\left( { - 1} \right)^{\,n} } \over 4}\left( { - 1} \right)^{\,n} n!\left( { - 3/4} \right)^{\underline {\, - 1 - n\,} } = {{\left( { - 1} \right)^{\,n + 1} n!} \over 4}\left( {3/4} \right)^{\overline {\, - n - 1\,} } = \cr & = {{\left( { - 1} \right)^{\,n + 1} } \over 4}{{\Gamma \left( {n + 1} \right)\Gamma \left( { - n - 1/4} \right)} \over {\Gamma \left( {3/4} \right)}} = {{\left( { - 1} \right)^{\,n + 1} } \over 4}{\rm B}\left( {n + 1,\; - n - 1/4} \right) = \cr & = {1 \over 4}{{n!} \over {\left( {1/4} \right)^{\overline {\,n + 1\,} } }} = {1 \over 4}{{\prod\limits_{k = 0}^{n - 1} {\left( {1 + k} \right)} } \over {\prod\limits_{k = 0}^n {\left( {1/4 + k} \right)} }} = \cr & = {{4^{\,n} } \over {\left( {1 + 4n} \right)}}\prod\limits_{k = 0}^{n - 1} {{{\left( {1 + k} \right)} \over {\left( {1 + 4k} \right)}}} = \cr & = {{4^{\,n} n!} \over {1 \cdot 5 \cdot 9 \cdot \; \ldots \; \cdot \left( {1 + 4n} \right)}} \cr} $$