Proving Binomial Identity without calculus 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  $$\sum_{0\le r\le n}\frac{(-1)^r\binom nr}{4r+1}=\frac{4^n\cdot  n!}{1\cdot5\cdot9\cdots(4n+1)}$$

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
$$
\eqalign{
  & S_{\,3} (n) = {{\left( { - 1} \right)^{\,n} } \over 4}\left. {\Delta ^{\,n} \left( {{1 \over {1/4 + x}}} \right)\,} \right|_{\,x\, = \,0}  =   \cr 
  &  = {{\left( { - 1} \right)^{\,n} } \over 4}\left( { - 1} \right)^{\,n} n!\left( { - 3/4} \right)^{\underline {\, - 1 - n\,} }
  = {{\left( { - 1} \right)^{\,n + 1} n!} \over 4}\left( {3/4} \right)^{\overline {\, - n - 1\,} }  =   \cr 
  &  = {{\left( { - 1} \right)^{\,n + 1} } \over 4}{{\Gamma \left( {n + 1} \right)\Gamma \left( { - n - 1/4} \right)} \over {\Gamma \left( {3/4} \right)}}
 = {{\left( { - 1} \right)^{\,n + 1} } \over 4}{\rm B}\left( {n + 1,\; - n - 1/4} \right) =   \cr 
  &  = {1 \over 4}{{n!} \over {\left( {1/4} \right)^{\overline {\,n + 1\,} } }}
 = {1 \over 4}{{\prod\limits_{k = 0}^{n - 1} {\left( {1 + k} \right)} } \over {\prod\limits_{k = 0}^n {\left( {1/4 + k} \right)} }} =   \cr 
  &  = {{4^{\,n} } \over {\left( {1 + 4n} \right)}}\prod\limits_{k = 0}^{n - 1} {{{\left( {1 + k} \right)} \over {\left( {1 + 4k} \right)}}}  =   \cr 
  &  = {{4^{\,n} n!} \over {1 \cdot 5 \cdot 9 \cdot \; \ldots \; \cdot \left( {1 + 4n} \right)}} \cr} 
$$
