sharp bounds for $(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{k}{n})$

Question

Is there any good bounds or estimation of $(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{k}{n})$ and $(1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{k}{n})$, $1 < k < n$?

What I actually want is $\sum_{1\le k \le n}(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{k}{n})$, and $\sum_{1\le k \le n}(1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{k}{n})$.

Answer 1

Classic products, whose results are $\left({\pm 1 \over n} \right)^k P(1\pm n,k)$ where $P$ is the Pochammer symbol.

And what you "really" want is:

$$-\frac{2^{2 n+1} e^{-n} n^{n+1} \Gamma \left(\frac{1}{2} (2 n+3)\right) \Gamma (-2 n-1,-n)}{\sqrt{\pi }}+e^{-n} (-n)^{n+1} \Gamma (-n,-n)-1$$

Answer 2

for the first part,

$$(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{k}{n})=y$$

Taking $\ln$ on both sides,

$$\ln(1+1/n)+\dots+\ln(1+k/n)=\ln(y)$$

Using the fact that $$ \ln(x+1 ) \le x$$

So $$\ln(y)\leq 1/n+\dots k/n=(1+2+3+\dots+k)/n=k(k+1)/2n$$

We get that $$y\leq e^{k(k+1)/2n}$$

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$$1-\frac{r(r-1)}{n} \le (1-\frac{1}{n})(1-\frac{2}{n})\cdots (1-\frac{r-1}{n})$$ Proof Am I correct when I do this in questions about induction?