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}

if we can write upper bound with $$ \sum_{n=0}^{N-1}(-1)^n\binom{N-1}{n}\frac{1}{n+cx} $$ is perfect7 — Mokhtar, Apr 28 '18 at 13:16
ok i hop if we can upper bound $$ \sum_{n=0}^{N-1}(-1)^n\binom{N-1}{n}\frac{1}{(n+1)*cx} $$ by $$ \sum_{n=0}^{N-1}(-1)^n\binom{N-1}{n}\frac{1}{n+cx} $$ — Mokhtar, Apr 28 '18 at 13:19
This is not clear. If that is your question, please post it in the question. — Simply Beautiful Art, Apr 28 '18 at 13:20
You cannot bound it like that so easily, since there is a $(-1)^n$ term in the sum. — Simply Beautiful Art, Apr 28 '18 at 13:25

score 2 · Accepted Answer · answered Apr 28 '18 at 13:28

2

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.

answered Apr 28 '18 at 13:28

Simply Beautiful Art

74,685

ok if there is any possibility that the term $\frac{1}{(cx)^N}$ appear in the upper bound – Mokhtar Apr 28 '18 at 13:35
what if we upper bound $$ \sum_{n=0}^{N-1}(-1)^n\binom{N-1}{n}\frac{1}{(n+1)cx} $$ by $$ \sum_{n=0}^{N-1}\binom{N-1}{n}\frac{1}{(n+1)cx} $$ since we have $cx$ is large so $1/((n+1)cx)$ will be close to zero $-1/((n+1)cx)$ also – Mokhtar Apr 28 '18 at 13:53
I see no reason for $1/(cx)^N$ to appear in relation to the sums. – Simply Beautiful Art Apr 28 '18 at 13:55
To the second comment, sure, that would be a valid bound, though I still see no point in the $cx$ in the denominator. – Simply Beautiful Art Apr 28 '18 at 13:56
ok sure i will try to do another think thank you so much – Mokhtar Apr 28 '18 at 14:04

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

1 Answers1