Upper and lower bound for fraction function : $ \sum_{n=0}^{N-1}(-1)^n\binom{N-1}{n}\frac{1}{(n+1)*cx} $

Question

Let $c$ and $N$ be two positive real number with $c>0$, and $x$ very high number. The number $c$ have two cases $c<1$ and $c\geq 1$.

if ther is any upper bound and lower bound for the following expression $$ \sum_{n=0}^{N-1}(-1)^n\binom{N-1}{n}\frac{1}{(n+1)*cx} $$ for the two case.

Can we upper bound $$ \sum_{n=0}^{N-1}(-1)^n\binom{N-1}{n}\frac{1}{(n+1)*cx} $$ by $$ \sum_{n=0}^{N-1}(-1)^n\binom{N-1}{n}\frac{1}{n+cx} $$ or can we do \begin{align} ncx+cx\geq& n+cx\\ \frac{1}{ncx+cx}\leq& \frac{1}{n+cx} \end{align}

Answer 1

Note that

$$\frac1{n+y}=\int_0^1x^{n+y-1}{\rm~d}x$$

and that

$$\sum_{n=0}^{N-1}\binom{N-1}n(-1)^nx^{n+y-1}=x^{y-1}(1-x)^{N-1}$$

which gives

$${\rm B}(y,N)=\int_0^1x^{y-1}(1-x)^{N-1}{\rm~d}x$$

which is the beta function. Assuming $y$ is a natural number, this gives us

$$\sum_{n=0}^{N-1}\binom{N-1}n\frac{(-1)^n}{n+y}=\frac{(y-1)!(N-1)!}{(y+N-1)!}$$

which reduces the problem to bounding the factorials, which can be done via things such as Stirling approximations.

And for the other sum,

$$\sum_{n=0}^{N-1}\binom{N-1}n\frac{(-1)^n}{(n+1)cx}=\frac1{Ncx}$$

which may be derived from the above.