Problem with a sequence

[um0123]

1. The problem statement, all variables and given/known data

Find n so that:

1/(1+√3) + 1/(√3+√5) + 1/(√5+√7) + .... + 1/(√2n-1 + √2n+1) = 100

Find n so that the same sum equals any number x (instead of 100)

2. Relevant equations

sum (1/(√2n-1 + √2n+1) = 100

3. The attempt at a solution

i have proven using the first 3 given terms that this series cannot be geometric or arithmatic. But if its neither how does my teacher expect me to find when the series equals 100?

what do i do? where do i go from here?

[Hurkyl]

Something.

You don't always have the luxury of knowing exactly what will work to solve a problem before actually trying it. Sometimes, you just have to try stuff to get ideas.

What have you tried? What might you consider trying? How have you solved other problems that may have vaguely resembled this one, or parts of this one?

[um0123]

i have been working with a friend thats in the same class for over 2 hours now and we dont really understand what exactly its asking for. I wrote the problem as it was written on the board in class, hes asking for the value of n so that the entire summation equals 100. but since the very first term isnt over 100 and it gets lower from there, will it never equals 100?

i wrote that i found its neither arithmatic or geometric so how can we solve it?

[Hurkyl]

Presumably he's asking for the value of n when that sum equals 100, not when the term equals 100.

The next question doesn't make sense, though.

[um0123]

So how do i find the value of n when the sum equals 100? hint please?

i wrote the entire thing as it appeared on the board, so if the second thing doesnt make sense its his fault :smile:

[Elucidus]

Try writing out the first few terms of the sequence of partial sums and see if there is a pattern. If there is a pattern, see if it leads to the n you are looking for. Note however that seeing a pattern does not mean it really exists. To fully "ice the cake" you need to prove the pattern is truly there (induction frequently is called for in for this), but if the question is only looking for n and no further proof is required, I don't think proving the existence of the pattern is necessarily expected here.

I believe the value of x in the second question is assumed to be "nice" in the sense that it is of a form that the series can assume (integer, rational, whatever). The pattern mentioned above should work. And in case you are wondering, a sum with continually diminishing terms can grow arbitrarily large.

--Elucidus

[PhaseShifter]

hint: {1\over{x+y}}={{x-y}\over{x^{2}-y^{2}}}

[um0123]

hint: {1\over{x+y}}={{x-y}\over{x^{2}-y^{2}}}

Thanks for the hint, im sure you meant it well. But i dont understand how that fits into my problem , you multiplied top and bottom by a giant 1 (x-y/x-y) to get the x-y/x^2 - y^2.

I have already tried multiplying the equation by (√2n-1 + √2n+1) so the summation is just (√2n-1 + √2n+1) instead of 1/(√2n-1 + √2n+1). BUT IT STILL DOESNT HELP ME!!!!!!!!!

am i just dumb?

[PhaseShifter]

You used a plus, I used a minus.

Try substituting x=\sqrt{2n-1} and y=\sqrt{2n+1} into the equation I gave. then see how you can simplify the results.

[Mark44]

The series looks like this in closed form:
\sum_{n = 1}^{\infty} \frac{1}{\sqrt{2n - 1} + \sqrt{2n + 1}}

You multiplied by \frac{\sqrt{2n - 1} + \sqrt{2n + 1}}{\sqrt{2n - 1} + \sqrt{2n + 1}}.
What if you multiply by 1 in the form \frac{\sqrt{2n - 1} - \sqrt{2n + 1}}{\sqrt{2n - 1} - \sqrt{2n + 1}}?

[um0123]

The series looks like this in closed form:
\sum_{n = 1}^{\infty} \frac{1}{\sqrt{2n - 1} + \sqrt{2n + 1}}

You multiplied by \frac{\sqrt{2n - 1} + \sqrt{2n + 1}}{\sqrt{2n - 1} + \sqrt{2n + 1}}.
What if you multiply by 1 in the form \frac{\sqrt{2n - 1} - \sqrt{2n + 1}}{\sqrt{2n - 1} - \sqrt{2n + 1}}?

if you multiplied it out would that be

(2n-1) - (√2n-1*√2n+1) + (√2n+1*√2n-1) - (2n+1)

im not sure how to multiply the middle parts. We havent learned complex numbers yet, and im pretty sure this includes it (but i might be wrong).

also, how do you make that big mth text, i would prefer to use it to make my writing more clear.

[lanedance]

click on the text to see the latex input, looks tricky but pretty simple once you've done it a few times
if you multiplied it out would that be

(2n-1) - (√2n-1*√2n+1) + (√2n+1*√2n-1) - (2n+1)

im not sure how to multiply the middle parts. We havent learned complex numbers yet, and im pretty sure this includes it (but i might be wrong).

also, how do you make that big mth text, i would prefer to use it to make my writing more clear.

this looks like you have multiplied out the denominator, note the "middle" parts cancel as they are exactly the negative of each other, leaving only (2n-1) - (2n+1).

For info, this multiplication doesn't have anything to do with complex numbers, they arise when you look at squareroots of negative numbers. As n>=1 in the sum we are only taking square roots of positive numbers

you should get to:
(\frac{1}{\sqrt{2n - 1} + \sqrt{2n + 1}}) (\frac{\sqrt{2n - 1} - \sqrt{2n + 1}}{\sqrt{2n - 1} - \sqrt{2n + 1}})

= \frac{\sqrt{2n - 1} - \sqrt{2n + 1}}{(2n-1) - (2n+1)}

then simplify & consider breaking it into 2 separate series and comparing the terms

[um0123]

click on the text to see the latex input, looks tricky but pretty simple once you've done it a few times

not sure how you get there multiplying, I get:
(\frac{1}{\sqrt{2n - 1} + \sqrt{2n + 1}}) (\frac{\sqrt{2n - 1} - \sqrt{2n + 1}}{\sqrt{2n - 1} - \sqrt{2n + 1}})

= \frac{\sqrt{2n - 1} - \sqrt{2n + 1}}{(2n-1) - (2n+1)}

then simplify & consider breaking it into 2 separate series and comparing the terms

ur top is correct, but when you multiply the bottom u need to dristribute the bottom out, like FOIL, if you know what that is. and i dont know if im right at all.

[lanedance]

edited my last post for clarity, so have another look & see if it helps

[PhaseShifter]

= \frac{\sqrt{2n - 1} - \sqrt{2n + 1}}{(2n-1) - (2n+1)}

then simplify & consider breaking it into 2 separate series and comparing the terms
Actually after you've simplified this see what happens when you add two consecutive terms for the series. Try n=x and n=x+1.

[um0123]

Actually after you've simplified this see what happens when you add two consecutive terms for the series. Try n=x and n=x+1.

okay, i simplify and get

√2n-1 - √2n+1
____________
4n

sorry i tried making the cool latex watever symbols but it didnt work out like i wanted it too. i need to do some research.

ANYWAY, this is as simplified as i can get it. I tried adding two consecutive numbers but it comes out to be some weird number. Am i doing something wrong?

[lanedance]

no quite there yet, for your denominator
(2n-1) - (2n+1) = 2n -1 - 2n -1 = -2

not 4n,

that should help with phaseshifter's suggestion, though will give you the same thing in the end as splitting into 2 series & shifting terms

[um0123]

no quite there yet, for your denominator
(2n-1) - (2n+1) = 2n -1 - 2n -1 = -2

not 4n,

that should help with phaseshifter's suggestion, though will give you the same thing in the end as splitting into 2 series & shifting terms

oh wow, sorry its pretty late and im kinda tired thats why i made the mistake of thinking it was 4n, this assignment is kicking my ***.

So its
√2n-1 - √2n+1
______________
-2

but i dont understand what you mean by splitting into 2 series and shifting terms.

[lanedance]

try phaseshifters suggestion first & add consecutive terms...

[um0123]

term 1:

√2(1)-1) - √2(1) + 1
__________________
-2

=

1 + √3
______
-2

term 2:

√2(2)-1) - √2(2) + 1
__________________
-2

=

√3 + √5
______
-2

adding those terms together gets

1 + √3 + √3 + √5
________________
-2


ARGH, I KNOW IM DOING SOMETHING WRONG.

[lanedance]

be careful with your negatives... getting close ;)

term 1:

(-1/2)(√(2(1)-1) - √(2(1) + 1))

= (-1/2)(√1 - √3)

[um0123]

term 1:

√2(1)-1) - √2(1) + 1
__________________
-2

=

1 - √3
______
-2

term 2:

√2(2)-1) - √2(2) + 1
__________________
-2

=

√3 - √5
______
-2

adding those terms together gets

1 - √3 + √3 + √5
________________
-2


1 + √5
______
-2

ARGH, I KNOW IM DOING SOMETHING WRONG.

i fixed my negatives in the quote and finished with the

1 - √5
______
-2

which i still dont see how that relates to my problem

edit: fixed the negative

[lanedance]

getting real close now

you missed another minus sign on the sqrt(5)

summing gives
1 term = (√3- 1)/2
2 terms = (√5- 1)/2
3 terms = ...

try adding one more term, then see if you can come up with a general pattern after summing N terms, then relate this to 100

[um0123]

getting real close now

you missed another minus sign on the sqrt(5)

summing gives
1 term = (√3- 1)/2
2 terms = (√5- 1)/2
3 terms = ...

try adding one more term, then see if you can come up with a general pattern after summing N terms, then relate this to 100

wait, what do you mean the first term is (√3- 1)/2, isnt it (1 - √3)/-2. and the second term is (1 - √5)/-2.

im sorry if im not getting something, but the terms you wrote are not what i got...and im sure that i have done something wrong.

EDIT: oh, you multiplied by a -1/-1, sorry. im really tired.

[Denyven]

I am in the same math class as Um0123, and I currently have got
\frac{\sqrt{2n-1} - \sqrt{2n+1}}{-2}

would squaring the fraction be a step in the right direction?

[um0123]

hey denny, yea this is the kid that i was working on this for 2 hours with, and it i showed him the physics forums and now hes trapped.

plus hes into partcile physics like me!:smile:

but back to busniess.

the 3rd term is

√5 - √7
_______
-2


add that to the previous summation of:

1 - √5
______
-2

and you get:

1 - √7
______
-2
now lets see that pattern. gimme a minute.

[PhaseShifter]

okay, i simplify and get

√2n-1 - √2n+1
____________
4n


Try again.

What is (\sqrt{2n-1})^{2}-(\sqrt{2n+1})^{2}?

[PhaseShifter]

I am in the same math class as Um0123, and I currently have got
\frac{\sqrt{2n-1} - \sqrt{2n+1}}{-2}

would squaring the fraction be a step in the right direction?

On the right track, but unfortunately you can't square this to get rid of the radicals for several reasons.
Try calculating that fraction for several consecutive values of n, and see if you notice a pattern.

[um0123]

is the pattern:

1 - √(n+2)
________
2

EDIT: DAMNIT, THAT ONLY WORKS FOR ODD TERMS!!!!

[lanedance]

no, try putting in the n values as a check, you will need to keep the squareroot

[lanedance]

notice it is always an odd term, if N is the number of terms and S is teh sum wth N terms:

N = 1,
S(N) = (√3- 1)/2

N = 2
S(N) = (√5- 1)/2

N = 3
S(N) = (√7- 1)/2
...
notice it is always includes squareroot of an odd term, any ideas how to write that in terms of N?

Once you have your general equation, solve it for N, when S(N) = 100


-----------------------------
I have to go now (good luck), but here's another more rigourous way to finish the problem:

to help see the pattern & summarise
S(N) = \sum_{n = 1}^{N} \frac{1}{\sqrt{2n - 1} + \sqrt{2n + 1}}

after rationalising the denominator
S(N) = \sum_{n = 1}^{N} \frac{1}{2} (\sqrt{2n + 1} - \sqrt{2n - 1} )

splitting the series & changing dummy to help with substitution:
S(N) = \frac{1}{2}( (\sum_{n = 1}^{N} \sqrt{2n + 1}) - (\sum_{m = 1}^{N}\sqrt{2m - 1}))

now to go forward form here:
- substitute into the 2nd sum only, m = n +1
- the sum limits of the 2nd sum will now become n = 0 to n = N-1 (previously m =1 to N)

cancelling terms should lead to the same pattern you observe

[Denyven]

Kk, I've done some work, and I think I found a pattern to the sum.

\frac{\sqrt{2n+1}-1}{2}

[um0123]

thanks lanedance, and phase shifter, and to everyone who helped.

together denny and i got the answer, and we couldnt ever do it without you guys.

[PhaseShifter]

Kk, I've done some work, and I think I found a pattern to the sum.

\frac{\sqrt{2n+1}-1}{2}

Right.

[Denyven]

SWEET, thanks guys, I got the answer. All your help, from everyone, was very much appreciated.