# How prove this inequality $\sum\limits_{k=1}^{n}\frac{1}{k!}-\frac{3}{2n}<\left(1+\frac{1}{n}\right)^n<\sum\limits_{k=1}^{n}\frac{1}{k!}$

show that $$\sum_{k=1}^{n}\dfrac{1}{k!}-\dfrac{3}{2n}<\left(1+\dfrac{1}{n}\right)^n<\sum_{k=0}^{n}\dfrac{1}{k!}(n\ge 3)$$

Mu try: I konw $$\sum_{k=0}^{\infty}\dfrac{1}{k!}=e$$

and I can prove Right hand inequality $$\left(1+\dfrac{1}{n}\right)^n=\sum_{k=0}^{n}\binom{n}{k}\dfrac{1}{n^k}$$ and note $$\binom{n}{k}\dfrac{1}{n^k}=\dfrac{1}{k!}\dfrac{(n-k+1)(n-k+2)\cdots(n-1)n}{n^k}\le\dfrac{1}{k!}$$

so $$\left(1+\dfrac{1}{n}\right)^n<\sum_{k=0}^{n}\dfrac{1}{k!}$$

But the left Hand inequality how prove it ? Thank you

• Also, lim n->infi (a+a/n)^n is e – LeeNeverGup Nov 11 '13 at 6:31

I think you mean this inequality$$\sum_{k=0}^{n}\dfrac{1}{k!}-\dfrac{3}{2n}<\left(1+\dfrac{1}{n}\right)^n.$$In fact, we can prove a sharper one $$\left(1+\frac{1}{n}\right)^n+\frac{3}{2n}>e.$$Let $f(x)=\left(1+\dfrac{1}{x}\right)^x+\dfrac{3}{2x}$, then$$f'(x)=\left(1+\dfrac{1}{x}\right)^x\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right)-\frac{3}{2x^2}.$$It's suffice to show that $f'(x)<0$ for $x>1$.Actually, notice that $\left(1+\dfrac{1}{x}\right)^x<3$ and $$\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\leqslant \frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}-\frac{1}{1+x}=\frac{1}{x^2(1+x)},$$thus for $x>1$, we have$$\left(1+\dfrac{1}{x}\right)^x\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right)-\frac{3}{2x^2}<\frac{3}{x^2(1+x)}-\frac{3}{2x^2}<0,$$therefore $f(x)$ is monotonically decreasing in $[1,+\infty)$. As $\displaystyle\lim_{x\rightarrow+\infty}f(x)=e$, we can get$$\left(1+\dfrac{1}{x}\right)^x+\dfrac{3}{2x}>e$$for all $x>1$, hence$$\left(1+\frac{1}{n}\right)^n+\frac{3}{2n}>e.$$