Turn on thread page Beta

problem watch

Announcements
    • Thread Starter
    Offline

    0
    ReputationRep:
    Consider the sequence an = ((n+1)0.5 ) - (n0.5 ) and bn = 1/(n0.5 )

    Let sn = a1 +a2 +a3 +...an = sum from k=1 to n ak and tn = b1 +b2 +b3 +...+bn = sum from k=1 to n bk be the corresponding sequences of so-called "partial sums".

    i) Determine the values of s5 , s20 and s100 . Hence or otherwise, simplify sn and explain why {sn } must tend to infinity.

    ii) Comfirm that an = 1 / (((n+1)0.5 )+(n0.5 )) and deduce that 0 < an < 1/2.bn for every n.
    Hence explain why tn > 2sn for every n and deduce that {tn } must also tend to infinity.

    :confused: :confused: :confused: :confused: :confused: :confused:
    Offline

    10
    ReputationRep:
    s_{n} = a_{n}+a_{n-1}+a_{n-2}+...a_{1} =\Big((n+1)^{\frac{1}{2}} - n^{\frac{1}{2}}\Big) + \Big(n^{\frac{1}{2}} - (n-1)^{\frac{1}{2}}\Big) + \Big((n-1)^{\frac{1}{2}} - (n-2)^{\frac{1}{2}}\Big) + ... + \Big(3^{\frac{1}{2}} - 2^{\frac{1}{2}}\Big) + \Big(2^{\frac{1}{2}} - 1^{\frac{1}{2}}\Big)

    See anything cancelling out? How can you compute the value of s_{n} for any n with ease?
    • Thread Starter
    Offline

    0
    ReputationRep:
    Ok I've been doing this for the majority of the weekend and still haven't progressed that much... I was thinking it was something to do with a arithmetic progression with a d of a1 is this right?
    • Thread Starter
    Offline

    0
    ReputationRep:
    :confused: :confused: :confused:
    Offline

    10
    ReputationRep:
    Look closely at what AlphaNumeric wrote:
    s_n = [sqrt(n+1) - sqrt(n)] + [sqrt(n) - sqrt(n-1)] + ... + [sqrt(3) - sqrt(2)] + [sqrt(2) - sqrt(1)]

    Perhaps re-bracketing it like this will make things clearer:
    s_n = sqrt(n+1) + [-sqrt(n) + sqrt(n)] + [-sqrt(n-1) + sqrt(n-1)] + ... + [-sqrt(3) + sqrt(3)] + [-sqrt(2) + sqrt(2)] - sqrt(1)

    Does you see what's going on?

    For part 2:
    a_n = sqrt(n+1) - sqrt(n) * [(sqrt(n+1) + sqrt(n))/(sqrt(n+1) + sqrt(n+1))]
    = [(sqrt(n+1)-sqrt(n))(sqrt(n+1)+sqrt(n))]/[sqrt(n+1) + sqrt(n)]
    = (n+1 - n)/[sqrt(n+1) + sqrt(n)]
    = 1/[sqrt(n+1) + sqrt(n)]

    a_n is obviously positive, and since n+1>n then 1/[sqrt(n+1) + sqrt(n)] < 1/[sqrt(n) + sqrt(n)] = 1/[2sqrt(n)] = (1/2)b_n.

    The last part should be easy now.
    • Thread Starter
    Offline

    0
    ReputationRep:
    Oooooh I see thanks you so much I can't believe the answer was staring me in the face the whole time
 
 
 

University open days

  • Manchester Metropolitan University
    Postgraduate Open Day Postgraduate
    Wed, 14 Nov '18
  • University of Chester
    Chester campuses Undergraduate
    Wed, 14 Nov '18
  • Anglia Ruskin University
    Ambitious, driven, developing your career & employability? Aspiring in your field, up-skilling after a career break? Then our Postgrad Open Evening is for you. Postgraduate
    Wed, 14 Nov '18
Poll
Should Banksy be put in prison?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Equations

Best calculators for A level Maths

Tips on which model to get

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.