Suppose $f(x)$ is continuous and decreasing on $[0,\infty]$, and $f(n) \to 0$. Define $\{a_n\}$ by $a_n = f(0)+f(1)+ \cdots + f(n-1) - \int_{0}^{n} f(x)dx$.
a). Prove $\{a_n\}$ is a Cauchy sequence directly from the definition.
b). Evaluate $\lim a_n$ if $f(x) = e^{-x}$.
For part a), I know that a sequence is Cauchy if it is bounded (below by 0?) which is given in the hypothesis (although I am not too sure why my book uses a closed bracket on $\infty$), if $\{a_n\}$ has a convergent subsequence $\{a_{n_i}\}$ which would follow from the Bolzano-Weierstass Theorem since $\{a_n\}$ is bounded, and lastly, if $\lim \{a_{n_i}\} \to L$ such that $\{a_n\} \to L$. This last part makes sense intuitively, but I cannot see how to show it in this particular problem. How does one pick this subsequence and ensure it is valid?
As for part b), I know that $\lim_{n\to\infty} f(x) =0$, but I believe it is asking for the sequence's limit, not the function. However, the integral should be equal to $e^{-n} - 1$ because of the minus sign in front of it. With all this in mind, wouldn't we still have all the terms going to 0?
I am using the textbook Introduction to Analysis by Arthur Mattuck. Thanks in advance; any help/advice/suggestions would be greatly appreciated.
To prove that the sequence is cauchy from the definition you can proceed as follows.
(a) Assume without loss of generality that $n\ge m$ $|a_{n}-a_{m}|=|(f(0)+...+f(n-1)-\int_{0}^{n}f(x)dx)-(f(0)+...+f(m-1)-\int_{0}^{m}f(x)dx)|=|(f(m)+...+f(n-1))+\int_{n}^{m}f(x)dx|$
$=|f(m)+...+f(n-1)-\int_{m}^{n}f(x)dx|=|f(m)+...+f(n-1)-\sum_{k=m}^{n-1}\int_{k}^{k+1}f(x)dx|\le|(f(m)+...+f(n-1))-(f(m+1)+...+f(n))|=|f(m)-f(n)|$
Since $\lim_{n\to\infty}f(n)=0$ then the sequence $\{f(n)\}_{n\in\mathbb{N}}$ is cauchy. Hence the above goes to $0$ as $n,m$ go to $\infty$. (Note that I used the fact that the function is decreasing to get the inequality.)
(b) If $f(x)=e^{-x}$ then $a_{n}=1+\frac{1}{e}+...+\frac{1}{e^{n-1}}-\int_{0}^{n}e^{-x}dx=\sum_{k=1}^{n}(\frac{1}{e})^{k-1}-(-e^{-x}|_{0}^{n})=\frac{1-(\frac{1}{e})^{n}}{1-\frac{1}{e}}-(1-e^{-n})$
$=\frac{1-(\frac{1}{e})^{n}}{1-\frac{1}{e}}+(\frac{1}{e})^{n}-1$.
As $n$ goes to $\infty$ then the above goes to:
$\frac{1}{1-\frac{1}{e}}-1=\frac{\frac{1}{e}}{1-\frac{1}{e}}=\frac{1}{e-1}$.