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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.

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  • $\begingroup$ Have you seen a proof of the integral test for series convergence? $\endgroup$ – Prahlad Vaidyanathan Sep 16 '13 at 2:19
  • $\begingroup$ Seen a proof of it, not quite. I am familiar with the integral test of series though. I believe that part b) of my question is clear to me. What is giving me some difficulty is the third condition to the Cauchy sequence part. That is, assuming that the first two conditions are also correct. $\endgroup$ – Jamil_V Sep 16 '13 at 2:24
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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:

$\frac{1}{1-\frac{1}{e}}-1=\frac{\frac{1}{e}}{1-\frac{1}{e}}=\frac{1}{e-1}$.

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  • $\begingroup$ In part a), the use of $\vert f(m)+ \cdots + f(n-1) + \int_{n}^{m} f(x) dx \vert$ is because of the assumption (without loss of generality) that $n \geq m$ right? Also, did the integral with the reversed bounds turn into a summation in order to show that the end result would be the same as the initial assumption of $ \vert a_n - a_m \vert$ ? Other than that, thank you very much, it is a very detailed post. $\endgroup$ – Jamil_V Sep 16 '13 at 2:36
  • $\begingroup$ Yes I used the assumption of $n≥m$ (without loss of generality) to get $|a_{n} −a_{m}|=|f(m)+...+f(n−1)+\int_{m}^{n}f(x)dx|$. I separated the integral into a sum because I wanted to estimate $|\sum_{k=m}^{n-1}f(k)-\int_{m}^{n}f(x)dx|$. I suspected that the end estimate would look like $|f(n)−f(m)|$ because I did not use the hypothesis $\lim_{n\to\infty}f(n)=0$ anywhere. You're very welcome. If you have any mroe questions I'd be happy to help. $\endgroup$ – user71352 Sep 16 '13 at 3:21

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