I have the following series at hand and I am tasked to rewrite it as an integral. This is what I've done so far. However I'm having trouble bringing my expression into the form of the given options.
Rewrite the given Binomial Series as an Integral.$$\dfrac{n\choose 0}{n(n+1)}-\dfrac{n\choose 1}{(n+1)(n+2)}+\cdots (n+1) \text{ terms}$$
1.) $\int_0^{1}x^{n-1}(1-x)^{n+1}\mathrm dx$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.)\int_{0}^{1}x^{n+1}(1-x)^{n-1}\mathrm dx$
My Attempt:
Clearly this would come from the Binomial expansion of $(1-x)^n$ with a term of $x^{\alpha}$ multiplied and has to be integrated twice. The $\alpha$ most likely seems to be $(n-1)$. And so I would write the series as $$\int_{0}^{1}\int_{0}^{x} x^{n-1}(1-x)^{n}\mathrm dx$$
How to bring this into the form as in the options?
Telescoping enables us to manage the calculation with one integral only. We obtain for $n\geq 1$: \begin{align*} \color{blue}{\sum_{j=0}^n}&\color{blue}{\binom{n}{j}(-1)^j\frac{1}{(n+j)(n+j+1)}}\\ &=\sum_{j=0}^n\binom{n}{j}(-1)^j\left(\frac{1}{n+j}-\frac{1}{n+j+1}\right)\\ &=\sum_{j=0}^n\binom{n}{j}(-1)^j\int_{0}^1\left(x^{n+j-1}-x^{n+j}\right)\,dx\\ &=\int_{0}^1\left(x^{n-1}-x^n\right)\sum_{j=0}^n\binom{n}{j}(-1)^jx^j\,dx\\ &=\int_0^1\left(x^{n-1}-x^n\right)(1-x)^n\,dx\\ &\,\,\color{blue}{=\int_0^1x^{n-1}(1-x)^{n+1}\,dx} \end{align*}
and the claim follows.