Error in approximating e via a finite sum

Question

I need some help with a homework problem.

I have to find an upper bound for the error in approximating $e$ by the series

$$\sum_{k=0}^{n} \frac{1}{k!}$$

I thought about using Taylor's theorem with the remainder term, but I didn't get very far with that.

Any suggestions?

Answer 1

You can avoid Taylor's theorem with some stingy estimations. Let $s_n = \sum_{k=1}^n \frac{1}{k!}$. Then \begin{align*} 0 < e - s_n &= \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \cdots \\ &= \frac{1}{(n+1)!}\left(1 + \frac{1}{(n+2)} + \frac{1}{(n+2)(n+3)} +\cdots \right) \\ &\leq \frac{1}{(n+1)!}\left(1 + \frac{1}{(n+1)} + \frac{1}{(n+1)^2} +\cdots \right) \\ &= \frac{1}{(n+1)!}\cdot \frac{n+1}{n} \\ &= \frac{1}{n!n}, \end{align*} where the equality in line 4 comes from geometric series.

Answer 2

For an upper bound,$$0<e-\sum_{k=0}^n k!^{-1}=\sum_{k=n+1}^{\infty} k!^{-1}=(n+1)!^{-1} \sum_{k=n+1}^{\infty}[ (n+1)! k!^{-1}]=$$ $$(n+1)!^{-1}\sum_{j=0}^{\infty} (n+1)!(n+1+j)!^{-1}<(n+1)!^{-1} \sum_{j=0}^{\infty}(n+1)^{-j}=$$ $$(n+1)!^{-1}((n+1)/n)=(n+1)!^{-1}(1+1/n).$$ A lower bound is $(n+1)!^{-1}$.