compute $\sum_{k=0}^{n} \frac{{(-1)^k}}{k+1}{n \choose k}$.

Question

Find $\sum_{k=0}^{n} \frac{{(-1)^k}}{k+1}{n \choose k}$ as a function of n.

I have done it in the following way:

Notice first that $\sum_{k=0}^{n} \frac{{(-1)^k}}{k+1}{n \choose k} = \sum_{k=0}^{\infty}\frac{(-1)^k}{k+1}{n \choose k}$ Because for $k > n$ we have $n \choose k$ = 0.

Now, using the binomial theorem:

$$(1+x)^n =\sum_{k=0}^{\infty}{n \choose k}x^k $$

$$(1-x)^n =\sum_{k=0}^{\infty}(-1)^k{n \choose k}x^k $$

$$\int(1-x)^ndx =\sum_{k=0}^{\infty}[(-1)^k{n \choose k}\int x^kdx] $$

$$ - \frac{(1-x)^{n+1}}{n+1} =\sum_{k=0}^{\infty}\frac{(-1)^k}{k+1}{n \choose k}x^{k+1}$$

Now let x = 1 and we get:

$$\sum_{k=0}^{\infty}\frac{(-1)^k}{k+1}{n \choose k} = - \frac{(1-1)^{n+1}}{n+1} = 0$$

However, the correct solution is $\frac{1}{n+1}$ and not $0$.

I cant find my mistake, can someone help me find it please? Thank you!

Edit: I have seen different ways to prove my question, but I would still like to know where is my mistake so i would know how to avoid it in the next time.

Answer 1

I think you forgot the constant term when integrating… $$\int(1-x)^ndx+C=\sum_{k=0}^{\infty}[(-1)^k{n \choose k}\int x^kdx] $$ Then let $x=0$ first, the correct number follows.

Answer 2

The problem is, when you integrated both sides, you forgot the constant of integration.

Instead of indefinite integration, you should apply the operator $\int_0^x$ to both sides. This does what you want on the right hand side, because $$ \int_0^x \chi^k d\chi= \frac {x^{k+1}}{k+1} $$ On the LHS, you will have $\int_0^x(1-\chi)^nd\chi$ which will provide the missing $1/(n+1)$.