Is the Sequence (1+\frac{1}{k})^{k+1} non-increasing for n>1?

[dobry_den]

Hey! I'm struggling with the proof of the following theorem:

Sequence $$(1+\frac{1}{k})^{k+1}$$, where k=1,2,3,..., is non-increasing.

Is there any chance of proving it by mathematical induction? Please could you give me a hint?

Thanks in advance!

 

[dobry_den]

Well, the proof doesn't have to be done by induction... I read this post (https://www.physicsforums.com/showthread.php?t=139672) and there might be some other solution, too.. Now I'm trying to express the sequence by the formula proposed by Werg22:

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

I hope it's possible... I'd be grateful for any hints:)

 

[Gib Z]

Non increasing means constant or decreasing. To show its not constant, just sub in 2 values of k, they won't be the same. To show its decreasing, find the derivative of the continuous version of the function and show that the derivative is always less than 0.

 

[sutupidmath]

well, i managed to prove it using a completely diffrent approach, if you are interested.?

 

[sutupidmath]

let us start from this: note this holds only for n>1
$$(1+\frac{1}{n^{2}-1})^{n} =(\frac{n^{2}-1+1}{(n-1)(n+1)})^{n} =(\frac{n}{n-1})^{n} (\frac{n}{n+1})^{n}$$ let us denote this by (1)

now using bernulis inequality we get

$$(1+\frac{1}{n^{2}-1})^{n}> 1+(\frac{n}{n^{2}-1}) > 1+(\frac{1}{n})$$---(2)

now we get

$$(\frac{n}{n-1})^{n} (\frac{n}{n+1})^{n} >1+(\frac{1}{n})$$
after we divide by $$(\frac{n}{n+1})^{n}$$ we get

$$(\frac{n}{n-1})^{n}> (1+ (\frac{1}{n}))^{n+1}$$
now let us do this little trick

$$(\frac{n-1+1}{n-1})^{n}=(1+ \frac{1}{n-1})^{n}>(1+ (\frac{1}{n}))^{n+1}$$

so now if we let $$b_(n-1)=(1+\frac{1}{n-1})^{n}$$ and $$b_n = (1+\frac{1}{n})^{n+1}$$ we see that b_(n-1)>b_n for every n>1