Proof of Cauchy Sequence for $\{a_n\}$ Defined by $f(x)$

[alexmahone]

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 $f(x)=e^{-x}$.

 

[tarantella]

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

 

[alexmahone]

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

Do you mean to interpret it geometrically as $T_1+T_2+\ldots+T_n$ as in the following figure?

https://www.physicsforums.com/attachments/36

 

[tarantella]

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]

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]

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

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

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

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

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

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

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

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

 

[CaptainBlack]

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