Let $u_n=\sum_{r=0}^n (-1)^r \frac{\binom nr}{x+r}$ at $x=2$. Then find the sum to infinity of $u_1+u_2..$

Question

Can I get a hint on how to begin the solution? I tried all possible derivative/integral methods of forming the general sum using the expansion of $(1+x)^n$, but I am just not able to the get $2+r$ in the denominator

Answer 1

We know that,

\begin{gather*} ( 1-x)^{n} =\sum ^{n}_{r=0}( -1)^{r} \cdot ^{n} C_{r} x^{r}\\ x( 1-x)^{n} =\sum ^{n}_{r=0}( -1)^{r} \cdot ^{n} C_{r} x^{r+1}\\ \int ^{1}_{0} x( 1-x)^{n} dx=\int ^{1}_{0}\sum ^{n}_{r=0}( -1)^{r} \cdot ^{n} C_{r} x^{r+1} dx\\ =\sum ^{n}_{r=0}\int ^{1}_{0}( -1)^{r} \cdot ^{n} C_{r} x^{r+1} dx\\ For\ LHS,\\ \int ^{1}_{0} x( 1-x)^{n} dx=\int ^{1}_{0}( 1-x) \cdot ( 1-( 1-x))^{n} dx\\ =\int ^{1}_{0} x^{n}( 1-x) dx=\frac{1}{n+1} -\frac{1}{n+2}\\ For\ RHS,\\ \sum ^{n}_{r=0}\int ^{1}_{0}( -1)^{r} \cdot ^{n} C_{r} x^{r+1} dx=\sum ^{n}_{r=0}( -1)^{r} \cdot \frac{^{n} C_{r}}{r+2}\\ So,\ as\ per\ your\ question,\ \\ u_{n} =\frac{1}{n+1} -\frac{1}{n+2}\\ for\ n=1\ to\ n=\infty . \end{gather*} Can you now complete the telescoping series and show that the result is 1/2?