Is this math problem solution wrong by any means?

[Nineshadow]

Problem itself :

X_n = 

 

Demonstrate that x_n < 1.

_______________________________________________________________

Solution : 

 

x_n= 

 

Now, is obviously smaller than 

So,applying this for all of them,we get :

 

x_n < 

 

x_n < 

 

x_n < 

 

 

Here comes the tricky one : 

Let's say I have 1/6.How do I write it as something minus something? 1/2 - 1/3.

Applying a logic similar to that,we get :

 =

 

Proof : ===

 

Thus:

x_n < 

 

So, then we get :

x_n  < 

x_n < 

Because 1/(2n+1) is subunitary , that means that 1 - 1/(2n+1) is smaller than 1.

Thus,  < 1/2 * 1 = 1/2 < 1 => x_n < 1

__________________________________________________________________________________________________________________________________________________________________

 

Anything wrong with it?

[aidenrelkoff]

TL;DR

 

naw, its look good.

[LukaP]

you know when you were taught multiplication and you had to "know it even if woken up at midnight"? well this is not like that looks good, but i CBA to do it myself now, way to late

[VMaxMuffin]

Here comes the tricky one :

Let's say I have 1/6.How do I write it as something minus something? 1/2 - 1/3.

Applying a logic similar to that,we get :

=

I think there's a mistake here. You're implying that:

(2k-1)(2k+1)=(1/2)(1/(2k-1) - 1/(2k+1))

which simply isn't true. Say k=3, then the left side would be 35 and the right would be 1/35. However it is correct to say

1/((2k-1)(2k+1))=(1/2)(1/(2k-1) - 1/(2k+1))

Hopefully you understand what I mean I'll come back and replace this with images later so it's easier to see.

[Nineshadow]

I think there's a mistake here. You're implying that:

(2k-1)(2k+1)=(1/2)(1/(2k-1) - 1/(2k+1))

which simply isn't true. Say k=3, then the left side would be 35 and the right would be 1/35. However it is correct to say

1/((2k-1)(2k+1))=(1/2)(1/(2k-1) - 1/(2k+1))

Hopefully you understand what I mean I'll come back and replace this with images later so it's easier to see.

Oh.Wait a sec.

So, if k = 1 :

1/3 = 1/2 * 2/3

1/3 = 1/3

That works.

 

Generally :

 

(2k+1 - 2k + 1)/(2k-1)(2k+1) = 2/...

1/2 * 2/... = 1/(2k-1)(2k+1) 

 

It's true.

[kingkickolas]

This line is fine:

=.

 

 

It's here that there's a problem:

x_n < 

going to

 

x_n  < .

 

You're missing the "1/(... - ...)" on all the terms.

[mansoor_]

its is correct, i have checked, you just mistype that one line(7th maths line). comparison test followed by the method of difference. 

[mansoor_]

 

 =   <---- RHS should not be reciprocal,

 

[VMaxMuffin]

It is 1/((2k-1)(2k-1)) though.

I know but on the other side you have

1/((1/2)(1/(2k-1) - 1/(2k+1))) when it should be just (1/2)(1/(2k-1) - 1/(2k+1))

So what I meant was that you said:

1/((2k-1)(2k-1)) = 1/((1/2)(1/(2k-1) - 1/(2k+1)))

When actually it's like this:

1/((2k-1)(2k-1)) = (1/2)(1/(2k-1) - 1/(2k+1))

[kingkickolas]

Oh yeah, actually

I think there's a mistake here. You're implying that:
(2k-1)(2k+1)=(1/2)(1/(2k-1) - 1/(2k+1))
which simply isn't true. Say k=3, then the left side would be 35 and the right would be 1/35. However it is correct to say
1/((2k-1)(2k+1))=(1/2)(1/(2k-1) - 1/(2k+1))

Hopefully you understand what I mean I'll come back and replace this with images later so it's easier to see.

 is totally right.

 

However, that makes the line

x_n  < 

correct,

which makes the end result correct.

[VMaxMuffin]

Oh yeah, actually

 is totally right.

 

However, that makes the line

x_n  < 

correct,

which makes the end result correct.

+1

Somehow I didn't notice that the rest of the solution followed correctly so it must have just been a typo.

[Nineshadow]

@VMaxMuffin @mansoor_

 

===

 

I made a small typo.Fixing it atm.

It should have been :

 

I actually wrote that,but I transcribed it wrong here.

[VMaxMuffin]

@VMaxMuffin

 

===

 

I made a small typo.Fixing it atm.

yes that's fine. I see now that it was a typo, if you fix that then the rest of your solution is perfectly fine.

[kingkickolas]

@VMaxMuffin @mansoor_

 

===

 

I made a small typo.Fixing it atm.

It should have been :

 

I actually wrote that,but I transcribed it wrong here.

Yes, that means that:

 

[Art Vandelay]

I dunno how mathematically sound it is to just state that the next is always smaller by giving an example with numbers.

 

If you wanted to be really sound, you could find the derivative of it, then find the point where the series reaches a maximum from that.

 

Alternatively, you could find the ratio of element n in the series to element n+1 in the series, and demonstrate that f(n + 1)/f(n) < 1 for all n > 0, to prove that it is decreasing.

[Nineshadow]

I dunno how mathematically sound it is to just state that the next is always smaller by giving an example with numbers.

 

If you wanted to be really sound, you could find the derivative of it, then find the point where the series reaches a maximum from that.

 

Alternatively, you could find the ratio of element n in the series to element n+1 in the series, and demonstrate that f(n + 1)/f(n) < 1 for all n > 0, to prove that it is decreasing.

If x < y < z, then obviously x <z right? D:

[Psykomantis00]

Yep... looks like math to me. ( I suck at math )

[VMaxMuffin]

 

x_n= 

 

Now, is obviously smaller than 

So,applying this for all of them,we get :

 

x_n < 

 

 

 

I dunno how mathematically sound it is to just state that the next is always smaller by giving an example with numbers.

 

If you wanted to be really sound, you could find the derivative of it, then find the point where the series reaches a maximum from that.

 

Alternatively, you could find the ratio of element n in the series to element n+1 in the series, and demonstrate that f(n + 1)/f(n) < 1 for all n > 0, to prove that it is decreasing.

 

Yes you're right, giving an example isn't really 'sufficient'. However in this case I don't think you'd be penalised (because it's kinda obvious). @Nineshadow In future you should just say:

 

4k² + 8k - 1 > 4k² - for all k>0

 

Therefore 1/(4k² + 8k - 1) < 1/(4k² - 1) for all k>0

 

Therefore x_n < Σ 1/(4k² -1)

[mr moose]

nope, you forgot to carry the 1.

[mansoor_]

I dunno how mathematically sound it is to just state that the next is always smaller by giving an example with numbers.

 

If you wanted to be really sound, you could find the derivative of it, then find the point where the series reaches a maximum from that.

 

Alternatively, you could find the ratio of element n in the series to element n+1 in the series, and demonstrate that f(n + 1)/f(n) < 1 for all n > 0, to prove that it is decreasing.

His method is perfectly fine.(comparison test &method of difference ).

Also finding the derivative does not tell you anything about the limit of the series, I think you meant integral of the function.

Also the ratio test you mentioned only tells you whether a series converges to a finite limit, not whether that limit is less than one. 

[Art Vandelay]

His method is perfectly fine.(comparison test &method of difference ).

Also finding the derivative does not tell you anything about the limit of the series, I think you meant integral of the function.

Also the ratio test you mentioned only tells you whether a series converges to a finite limit, not whether that limit is less than one. 

I dunno, if I handed something in with a proof like that, at my uni, they'd give me a 0.

A critical point exists where the first derivative equals 0. If you can show the first derivative is always negative on x > 0, the function is always decreasing, since the first derivative is the rate of change.

You don't take the limit, I meant like you just analyse f(n+1)/f(n). Take the harmonic series, for example, f(n+1) = 1/(x+1) and f(n) = 1/x, f(n+1)/f(n) = x/(x+1). Thus, you've shown that the ratio will always be less than 1 (for n > 0). The limit of this, however is 1, so it isn't convergent. I didn't put too much thought into that, but it seems right.

 

Also, I had a prof named Mansour once. O_o

 

 

Yes you're right, giving an example isn't really 'sufficient'. However in this case I don't think you'd be penalised (because it's kinda obvious).

Depends what kind of course you're taking. University level courses will give you a 0 for that kind of thing.

 

 

If x < y < z, then obviously x <z right? D:

You have proven that x < y, but you haven't proven y < z for all numbers within the domain of this question.

[mansoor_]

 

 

Sorry i can only screen shot, converting between latex and html is difficult. 

[Art Vandelay]

 

Sorry i can only screen shot, converting between latex and html is difficult. 

I think I just misread the question and answer horribly. I feel stupid now.

 

For some reason I thought he was trying to implicitly use recursion to solve that, and I also missed the sum symbol in the question.

 

I'm going to stop answering questions online from now on.

[Nineshadow]

I dunno, if I handed something in with a proof like that, at my uni, they'd give me a 0.

A critical point exists where the first derivative equals 0. If you can show the first derivative is always negative on x > 0, the function is always decreasing, since the first derivative is the rate of change.

You don't take the limit, I meant like you just analyse f(n+1)/f(n). Take the harmonic series, for example, f(n+1) = 1/(x+1) and f(n) = 1/x, f(n+1)/f(n) = x/(x+1). Thus, you've shown that the ratio will always be less than 1 (for n > 0). The limit of this, however is 1, so it isn't convergent. I didn't put too much thought into that, but it seems right.

 

Also, I had a prof named Mansour once. O_o

Well...I'm in the 9th grade.Still a long way to uni,

[WolfDeville]

uhm... 5