0
$\begingroup$

Let $R(z)$ be the remainder after $n$ terms in the power series of $e^z$. That is $$R(z) = e^z - \sum_{k=1}^{n}\frac{z^k}{k!}=\sum_{k=n+1}^{\infty}\frac{z^k}{k!}$$ Prove that $|R(z)| \leq \frac{e-1}{(n+1)!}$ if $|z| \leq 1$

I have been expanding the series trying to see if I can find an inequality but had a hard time going back to a series that would give me $e-1$. I'm thinking it requires integrals to prove it.

$\endgroup$
1
  • $\begingroup$ The power series for $e^z$ should start with$k=0$ not with $k=1$. The $k$th term is $z^{k-1}/(k-1)!$, not $z^k/k!$. $\endgroup$ Apr 4, 2016 at 0:55

1 Answer 1

0
$\begingroup$

$|z|\leq 1\implies$ $|\sum_{j=1}^{\infty}\frac {z^{n+j}}{(n+j)!}|\leq \sum_{j=1}^{\infty}\frac {1}{(n+j)!}=$ $$= \frac {1}{(n+1)!}(1+\frac {1}{(n+2)}+\frac{1}{(n+2))(n+3)} +\frac {1}{(n+2)(n+3)(n+4)}+...)\leq$$ $$\leq \frac {1}{(n+1)!}(1+\frac {1}{2}+\frac {1}{2\cdot 3}+\frac {1}{2\cdot 3 \cdot 4}+...)=\frac {1}{(n+1)!}(e-1).$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .