You are Here: Home >< Maths

# problem watch

Announcements
1. 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.

2. See anything cancelling out? How can you compute the value of for any n with ease?
3. 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?
4. 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.
5. Oooooh I see thanks you so much I can't believe the answer was staring me in the face the whole time

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: November 21, 2005
Today on TSR

Choose Respect

### University open days

• Manchester Metropolitan University
Wed, 14 Nov '18
• University of Chester
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
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams