Question: Can the Decreasing Sequence be Proven Using Different Methods?

[murshid_islam]

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

this is what i did:

$$a_k = {\left(\frac{k}{k+1}}\right)^k$$

$$a_{k+1}-a_{k}$$

$$= {\left(\frac{k+1}{k+2}}\right)^{k+1}-{\left(\frac{k}{k+1}}\right)}^{k}$$

$$= {\left(\frac{1+\frac{1}{k}}{1+\frac{2}{k}}}\right)^{k+1}-{\left(\frac{k}{k+1}}\right)^{k}$$

$$< {\left(\frac{1+\frac{1}{k}}{1+\frac{1}{k}}}\right)^{k+1}-{\left(\frac{k}{k+1}}\right)^{k}$$

$$= 1 - \left({\frac{k}{k+1}}\right)^{k}$$

$$< 1-1$$ since $$\left({\frac{k}{k+1}}\right)^{k} < 1$$

therefore, $$a_{k+1}-a_{k} < 0$$

therefore, the sequence is decreasing.

am i right?

 

[arildno]

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

 

[Office_Shredder]

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

a_k+1 - a_k = 0 < 1.

Therefore, the sequence is decreasing :P

 

[murshid_islam]

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?

 

[arildno]

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.

 

[Werg22]

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 (unfortunately), 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

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

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

$$S_n = 1 + n\frac{1}{n} + \frac{1}{2!}(1-\frac{1}{n})+ ... + \frac{1}{n!}(1 - \frac{1}{n})(1 - \frac{2}{n})...(1 - \frac{n-1}{n})$$,

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.

 

[murshid_islam]

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 $$y = \ln\left(1+\frac{1}{x}\right)$$ then i get $$y' = \frac{-1}{x(x+1)}$$

 

[murshid_islam]

is there any way to prove that the sequence is decreasing the way i started?

 

[Werg22]

is it correct?
if $$y = \ln\left(1+\frac{1}{x}\right)$$ then i get $$y' = \frac{-1}{x(x+1)}$$

Wait, wait, wait, my proof is erroneous. $$y = \ln\left(1+\frac{1}{x}\right)$$ then $$y' = \frac{-1}{x(x+1)}$$

You are correct. I'll edit my proof right now.

 

[murshid_islam]

is there any way to prove the sequesnce is decreasing by one of the following:
$$a_{k+1}-a_k < 0$$

$$\frac{a_{k+1}}{a_k} < 1$$

$$f'(k) < 0$$