Decreasing sequence (proof)

i have to prove that the sequence {ak} is decreasing, where
[tex]\{a_k\} = \{\left({1+\frac{1}{k}}\right)^{-k}\}[/tex]

this is what i did:

[tex]a_k = {\left(\frac{k}{k+1}}\right)^k[/tex]

[tex]a_{k+1}-a_{k}[/tex]

[tex]= {\left(\frac{k+1}{k+2}}\right)^{k+1}-{\left(\frac{k}{k+1}}\right)}^{k}[/tex]

[tex]= {\left(\frac{1+\frac{1}{k}}{1+\frac{2}{k}}}\right)^{k+1}-{\left(\frac{k}{k+1}}\right)^{k}[/tex]

[tex]< {\left(\frac{1+\frac{1}{k}}{1+\frac{1}{k}}}\right)^{k+1}-{\left(\frac{k}{k+1}}\right)^{k}[/tex]

[tex]= 1 - \left({\frac{k}{k+1}}\right)^{k}[/tex]

[tex]< 1-1[/tex] since [tex]\left({\frac{k}{k+1}}\right)^{k} < 1[/tex]

therefore, [tex]a_{k+1}-a_{k} < 0[/tex]

therefore, the sequence is decreasing.

am i right?
 
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arildno

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Hmm..no
The RATIO between the terms should be less than 1, not the DIFFERENCE!
The difference should be less than 0
 

Office_Shredder

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Suppose we take the sequence 1/2, 1/2, 1/2.

ak+1 - ak = 0 < 1.

Therefore, the sequence is decreasing :P
 
arildno said:
Hmm..no
The RATIO between the terms should be less than 1, not the DIFFERENCE!
The difference should be less than 0
ok, i made a stupid mistake. i have edited my proof. is it correct now?
 
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arildno

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No, it is not.
The reason is that since the last fraction is LESS than 1, the DIFFERENCE between 1 and the fraction is GREATER than the difference between 1 and 1.
Your first bound is too crude to derive the result.
 
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I suggest the following:

Put the expression under the form of e^f(k). The derivative of that expression would be e^f(k) * f'(k). Knowing that e^f(k) is positive for k sufficiently large, find the sign that f'(k) assumes for an also sufficiently k. If the sign is negative, the expression e^f(k) * f'(k) is negative for this value of k and beyond, and you have proven that e^f(k) is decreasing.

Edit: you will find f(k) = ln (1 + 1/k) * - k. If y = ln (1 + 1/k), y' = -1/k(k+1) and f'(k) = -ky' - y = 1/(k+1) - ln (1 + 1/k).

Now (unfortunatly), this is not relevant. We are confronted to another inequality, ln (1 + 1/k) > 1/(k+1).

Sorry that I've wasted so much time... maybe the second inequality would help, but I doubt it. To compensate I'll give you another proof , we have:

(1+1/k)^k is increasing, for

[tex] (1+1/n)^{n} = 1 + n\frac{1}{n} + \frac{1}{2!}(1-\frac{1}{n})+ ... + \frac{1}{n!}(1 - \frac{1}{n})(1 - \frac{2}{n})...(1 - \frac{n-1}{n}) [/tex].

This is simply by using the binominial theorem. Now if,

[tex]S_n = 1 + n\frac{1}{n} + \frac{1}{2!}(1-\frac{1}{n})+ ... + \frac{1}{n!}(1 - \frac{1}{n})(1 - \frac{2}{n})...(1 - \frac{n-1}{n}) [/tex],

then S_m > S_n. Since (1+1/m)^m = S_m + ...., then (1+1/m)^m > (1+1/n)^n

and

(1+1/m)^-m < (1+1/n)^-n

Thus the expression is decreasing. It should be noted that this is only a proof for integrer values of n.
 
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Werg22 said:
you will find f(k) = - ln (1 + 1/k) * k. You have y = ln (1 + 1/k), y' = ln k/(1 + 1/k)
is it correct?
if [tex]y = \ln\left(1+\frac{1}{x}\right)[/tex] then i get [tex]y' = \frac{-1}{x(x+1)}[/tex]
 
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is there any way to prove that the sequence is decreasing the way i started?
 
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murshid_islam said:
is it correct?
if [tex]y = \ln\left(1+\frac{1}{x}\right)[/tex] then i get [tex]y' = \frac{-1}{x(x+1)}[/tex]
Wait, wait, wait, my proof is erroneous. [tex]y = \ln\left(1+\frac{1}{x}\right)[/tex] then [tex]y' = \frac{-1}{x(x+1)}[/tex]

You are correct. I'll edit my proof right now.
 
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is there any way to prove the sequesnce is decreasing by one of the following:
[tex]a_{k+1}-a_k < 0[/tex]

[tex]\frac{a_{k+1}}{a_k} < 1[/tex]

[tex]f'(k) < 0[/tex]
 

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