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#### Alexmahone

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- Jan 26, 2012

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$a_n=f(0)+f(1)+\ldots+f(n-1)-\int_0^n f(x)dx$

(a) Prove $\{a_n\}$ is a Cauchy sequence directly from the definition.

(b) Evaluate $\lim a_n$ if $f(x)=e^{-x}$.

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- #1

- Jan 26, 2012

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$a_n=f(0)+f(1)+\ldots+f(n-1)-\int_0^n f(x)dx$

(a) Prove $\{a_n\}$ is a Cauchy sequence directly from the definition.

(b) Evaluate $\lim a_n$ if $f(x)=e^{-x}$.

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- Jan 26, 2012

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Do you mean to interpret it geometrically as $T_1+T_2+\ldots+T_n$ as in the following figure?We can write $a_n=\sum_{k=0}^{n-1}\int_k^{k+1}(f(k)-f(x))dx$. Does it help you?

View attachment 36

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- Jan 26, 2012

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I think I can show it geometrically.Now you have to show analytically that $\{a_n\}$ is a Cauchy sequence.

$a_m-a_n=T_{n+1}+T_{n+2}+\ldots+T_m$ for $m>n$

Given $\epsilon>0$,

$f(n+1)<\epsilon$ for $n\gg 1$

By moving all the "triangles" from $T_{n+1}$ to $T_m$ horizontally to the left into the rectangle of base 1 and height $f(n+1)$ (as shown in the figure), we see that

$T_{n+1}+T_{n+2}+\ldots+T_m<f(n+1)<\epsilon$

Does that look ok?

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- Jan 26, 2012

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(b) $a_n=\sum_{k=0}^{n-1}\int_k^{k+1}(f(k)-f(x))dx$

$=\sum_{k=0}^{n-1}\int_k^{k+1}(e^{-k}-e^{-x})dx$

$=\sum_{k=0}^{n-1}\int_k^{k+1}(xe^{-k}+e^{-x})$

$=\sum_{k=0}^{n-1}\int_k^{k+1}((k+1)e^{-k}+e^{-k-1}-ke^{-k}-e^{-k})$

$=\sum_{k=0}^{n-1}e^{-k-1}$

$\lim a_n=\sum_{k=0}^\infty e^{-k-1}$

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

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

$=\sum_{k=0}^{n-1}\int_k^{k+1}(e^{-k}-e^{-x})dx$

$=\sum_{k=0}^{n-1}\int_k^{k+1}(xe^{-k}+e^{-x})$

$=\sum_{k=0}^{n-1}\int_k^{k+1}((k+1)e^{-k}+e^{-k-1}-ke^{-k}-e^{-k})$

$=\sum_{k=0}^{n-1}e^{-k-1}$

$\lim a_n=\sum_{k=0}^\infty e^{-k-1}$

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

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

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- Jan 26, 2012

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The first thing we need is that for a non-negative decreasing function:

$a_n=f(0)+f(1)+\ldots+f(n-1)-\int_0^n f(x)dx$

(a) Prove $\{a_n\}$ is a Cauchy sequence directly from the definition.

(b) Evaluate $\lim a_n$ if $f(x)=e^{-x}$.

\[ f(k+1) \le \int_k^{k+1} f(x)\;dx \le f(k) \]

Hence for \(m>n\):

\[ \sum_{k=n+1}^{m}f(k) \le \int_n^m f(x)\;dx\le \sum_{k=n}^{m-1}f(k) \]

Now :

\[ a_m-a_n=\sum_{k=n}^{m-1}f(k) -\int_n^mf(x)\;dx \]

So:

\[ \sum_{k=n}^{m-1}f(k) -\sum_{k=n}^{m-1}f(k) \le a_m-a_n \le \sum_{k=n}^{m-1}f(k) -\sum_{k=n+1}^{m}f(k) \]

simplifying:

\[ 0 \le a_m-a_n \le f(n)-f(m) \]

Hence \( \displaystyle \lim_{n,m\to \infty}|a_m-a_n|=0 \)

CB

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