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Cauchy sequence

Alexmahone

Active member
Jan 26, 2012
268
Suppose $f(x)$ is continuous and decreasing on $[0, \infty]$, and $f(n)\to 0$. Define $\{a_n\}$ by

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

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

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

girdav

Member
Feb 1, 2012
96
We can write $a_n=\sum_{k=0}^{n-1}\int_k^{k+1}(f(k)-f(x))dx$. Does it help you?
 

Alexmahone

Active member
Jan 26, 2012
268

girdav

Member
Feb 1, 2012
96
Yes, it will give you the idea, and we can see what $a_n$ represents. Now you have to show analytically that $\{a_n\}$ is a Cauchy sequence.
 

Alexmahone

Active member
Jan 26, 2012
268
Now you have to show analytically that $\{a_n\}$ is a Cauchy sequence.
I think I can show it geometrically.

$a_m-a_n=T_{n+1}+T_{n+2}+\ldots+T_m$ 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?
 

Alexmahone

Active member
Jan 26, 2012
268
(b) $a_n=\sum_{k=0}^{n-1}\int_k^{k+1}(f(k)-f(x))dx$

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

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

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

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

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

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

$=\frac{1}{e-1}$
 
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CaptainBlack

Well-known member
Jan 26, 2012
890
Suppose $f(x)$ is continuous and decreasing on $[0, \infty]$, and $f(n)\to 0$. Define $\{a_n\}$ by

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

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

(b) Evaluate $\lim a_n$ if $f(x)=e^{-x}$.
The first thing we need is that for a non-negative decreasing function:

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

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

You will need to check that the above is correct, as it is too easy for the odd index to go wrong here and there, as it has done numerous times while constructing this post (Angry)

CB
 
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