Got a tricky question! (1 Viewer)

[adzy]

A question 8 from a trial paper

Evaluate summation from n=1 to infinity of 1/(1+u_n) given that (u_n+1) = u_n + (u_n)^2 and u_1 = 1/3

I go to as far as summation of n=1 to infinity of: U_n / U_n+1 but uh...I don't think that really does anything.

 

[justchillin]

for this question to be answered im assuming they've given u a lead-in...ie some other expression for u_n. If this is the case, which I'm sure it is, then the limit as n approaches infinity is equal to (theoritically) the limit as n approaches infinity plus one. Let the limit = M say and evaluate (proably a quadratic in M). Hope that helps...

 

[MAICHI]

u_n actually goes off to infinity or else the summation can't converge. The series converges by comparing to geometric series R^n with R<1, because the ratio of every next term is smaller than 1. But I don't see any simple way you can find an expression for R, maybe some other method.

 

[KFunk]

Which trial is the question from?

 

[damo676767]

by u_n do you meen u_n ?

 

[MAICHI]

by u_n do you meen u_n ?

I think it's like this.

*sigh*, anyone got any clues? I kinda wish the series actually diverges, so you just write infinity.

 

[Stefano]

I think it's like this.

*sigh*, anyone got any clues? I kinda wish the series actually diverges, so you just write infinity.

Is that 4 unit? My school never covered sums to infinity.

 

[acmilan]

I did a short C code and, unless i made an error, it seems to diverge. So maybe you just need to prove it diverges
Edit: no it doesnt, my mistake, it seems to converge to 3.

#include < stdio.h>

int main()
{
    float sum = 0, i = 0, u_n = 1.0/3.0;
    for (i; i < 10; i++)
    {
        sum = sum + 1/(1 + u_n);
        u_n = u_n + u_n*u_n;
        printf("%8f\n", sum);
    }
}

Output:
0.750000
1.442308
2.051330
2.538160
2.854090
2.981421
2.999661
3.000000
3.000000
3.000000

 

[KFunk]

I'd like to see the lead in to the question.

 

[justchillin]

I dont think you can do this without a lead in... its like finding the limit to infinity of the fibonnaci series say... find 2 experssions for the same thing, let lim n app infinity = lim n app infinity + 1...then etc

 

[adzy]

This is SGS 1994 4U Trial Paper

There is no follow up question but the preevious questions before that were:

a) If I_n = integral x^n * e^(x^2) dx show that I_n + (n-1) I_(n-2) = 2e (n>=2)

Edit: Shoulw the answer read I_n + (n-1)/2 (or times two?) * I_(n-2) = 2e? What they have is what I typed above but my answer to that doesn't seem to match.

ii) Evaluate I_5

b) f(x) is given by f(x) = x - ln (1+x^2)
i) Show that f'(x) >=0 for all values of x
ii) Deduce that e^x > 1+x^2 for all positive values of x
c) The summation question

There doesn't seem to be a lead in question and I'm not sure that the previous questions are direclty related.

Yeah, the question is exactly as MAICHI's picture says.

 

[justchillin]

U_n+1 = U_n(1+U_n)
Let u_n = n (for my typing sake)
1/n - 1/(N+1) = 1/n - 1/n(1+n)
= 1+ n -1/ (n(1+n))
1/ 1+n
now.....
sum to infinity of 1/1+n = 1/1+U-1+ 1/1+u_2+.....
= (1/u_1 - 1/u_2) + (1/u_2 - 1/U-3) +.....
= 1/U_1
= 1/(1/3)
= 3

 

[Stefano]

Could this stuff be examined? I've never seen it before!

 

[justchillin]

Yeh it comes under harder 3 unit...this question doesnt do anything new anyway, u just have to play around with the expression...

 

[KFunk]

U_n+1 = U_n(1+U_n)
Let u_n = n (for my typing sake)
1/n - 1/(N+1) = 1/n - 1/n(1+n)
= 1+ n -1/ (n(1+n))
1/ 1+n
now.....
sum to infinity of 1/1+n = 1/1+U-1+ 1/1+u_2+.....
= (1/u_1 - 1/u_2) + (1/u_2 - 1/U-3) +.....
= 1/U_1
= 1/(1/3)
= 3

Nice method, it's very simple. I geuss you can safely assume that u_n --> &infin; as n--> &infin; but first would you show that there is a term u_k where u_k > 1 or just not bother?

 

[MAICHI]

Well done justchillin, *pat on back*.

Thats called a telescoping series method what justchillin just used. But that's a bad question, unless telescoping series is a popular question in these type of exams, which is really not, you should really be given a lead in.

 

[justchillin]

I dont think you'd have 2 bother? But im a lazy man... And yes telescoping series is exactly what I used...