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

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2 Answers 2

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.

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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

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