Sum of $\frac{C_r}{(r+1)(r+2)}$ from $r=0$ to $r=n$

Question

I am trying to compute the following sum: $$\sum_{r=0}^{n}\frac{C_r}{(r+1)(r+2)}$$

Integrating $(1+x)^n = \sum_{r=0}^{n}C_rx^r$ with respect to $x$, $$\int(1+x)^n dx=\int\sum_{r=0}^{n}C_rx^rdx$$ $$\implies \frac{(1+x)^{n+1}}{n+1}=\sum_{r=0}^{n}C_r\frac{x^{r+1}}{r+1}$$

Again integrating with respect to $x$ from $0$ to $1$ yields, $$\int_0^1\frac{(1+x)^{n+1}}{n+1}dx=\int_0^1\sum_{r=0}^{n}C_r\frac{x^{r+1}}{r+1}dx$$

$$\implies\bigg[\frac{(1+x)^{n+2}}{(n+1)(n+2)}\bigg]_0^1=\bigg[\sum_{r=0}^{n}C_r\frac{x^{r+2}}{(r+1)(r+2)}\bigg]_0^1$$

$$\implies\sum_{r=0}^{n}\frac{C_r}{(r+1)(r+2)}=\frac{2^{n+2}-1}{(n+1)(n+2)}$$

But it is written in my textbook that the answer is $$\frac{2^{n+2}-(n+3)}{(n+1)(n+2)}$$

So, where am I going wrong? Is it correct to use integral twice in such situations?

Answer 1

We have to respect the integration constant when evaluating an indefinite integral.

When calculating $\int(1+x)^n\,dx = \int\sum_{r=0}^nC_rx^r\,dx$ we obtain \begin{align*} \int(1+x)^n\,dx&=\frac{(1+x)^{n+1}}{n+1}+K_1\\ \int\sum_{r=0}^nC_rx^r\,dx&=\sum_{r=0}^nC_r\frac{x^{r+1}}{r+1}+K_2 \end{align*} with $K_1,K_2$ constants.

It follows \begin{align*} \frac{(1+x)^{n+1}}{n+1}=\sum_{r=0}^nC_r\frac{x^{r+1}}{r+1}\color{blue}{+K}\tag{1} \end{align*} with $K=K_2-K_1$.

By setting $x=0$ in (1) we obtain \begin{align*} \color{blue}{\frac{1}{n+1}=K} \end{align*}

We conclude \begin{align*} \frac{(1+x)^{n+1}}{n+1}=\sum_{r=0}^nC_r\frac{x^{r+1}}{r+1}\color{blue}{+\frac{1}{n+1}}\tag{2} \end{align*}

Integrating (2) from $0$ to $1$ gives now the correct result

\begin{align*} \color{blue}{\int_{0}^1\sum_{r=0}^nC_r\frac{x^{r+1}}{r+1}\,dx}&=\int_{0}^1\left(\frac{(1+x)^{n+1}}{n+1}-\frac{1}{n+1}\right)\,dx\\ &=\left[\frac{(1+x)^{n+2}}{(n+1)(n+2)}-\frac{x}{n+1}\right]_0^1\\ &=\left[\frac{2^{n+2}}{(n+1)(n+2)}-\frac{1}{n+1}\right]-\left[\frac{1}{(n+1)(n+2)}\right]\\ &\color{blue}{=\frac{2^{n+2}-(n+3)}{(n+1)(n+2)}} \end{align*} as expected.

Answer 2

I think your answer is correct because also we have: $$\sum_{r=0}^n\frac{\binom{n}{r}}{(r+1)(r+2)}=\frac{1}{(n+1)(n+2)}\sum_{r=0}^n\binom{n+2}{r}=\frac{1}{(n+1)(n+2)}\left(2^{n+2}-1-(n+2)\right)$$