Very hard sequences question!!?

Question

the sequece {tn} is given by

tn = 1/(n+1) + 1/(n+2) + 1/(n+3) + ........+ 1/(4n).

Show that tn is monotone, and that it converges to a limit L where

13/12 < L ≤ 3.

Cheers

Answer 1

For every n,

t(n) = 1/(n+1) + 1/(n+2) + 1/(n+3) + ........+ 1/(4n) and
t(n +1) = 1/(n+2) + 1/(n+3) + ........+ 1/(4(n+1)), so that

t(n +1) - t(n) = 1/(4n +1) + ....1/(4n+4)) - 1/(n +1) > 4/(4n +4) - 1/(n +1) = 1/(n+1) - 1/(n +1) = 0, Therefore t(n +1) - t(n) >0 for every n, which shows t(n) is strictly monotone increasing.

For every n, t(n) = 1/(n+1) + 1/(n+2) + 1/(n+3) + ........+ 1/(4n) < 3n/(n +1) . Since 3n/(n +1) -> 3 as n -> oo, if follows t(n) <=3 for every n. This shows t(n) is bounded and converges to supremum t(n). So, t(n) converges to a limit L <=3.

On the other hand, t(1) = 1/2 + 1/3 +1/4 = (6 + 4 + 3)/12 = 13/12. Since t(n) is strictly montone increasing, for n>=3 we have t(n) > t(2) >t(1) =13/12, which implies limt(n) = L >= t(2) > t(1) = 13/12 => L >13/12.

Therefore, we have 13/12 < L <= 3.

Answer 2

If by monotone you mean continually increasing then that is fairly easy .

Consider t(n) and t(n +1)

t(n+1) - t(n) =1/(4n+1) +1/(4n+2) +1/(4n+3)+1/(4n+4)-1/(n+1)

Note that if you sum the first 4 terms this will always be greater than 4/(4n +4) =1/(n+1) . So therefore t(n+1) >t(n)


QED

It seems to me also that the limit of the sequence as n goes to infinity is always going to be less than 3n/n+1 . This limit = 3.

The reason for this is that there are going to be 3n terms of 1/(n +1) or less.

The same method would put a lower limit of 3/4 i.e. 3n terms of 1/4n or more but haven't yet got to 13/12.

Addendum to answer

To get the 13/12 answer you spilt the sequence into three sections. from 1/(n +1) to 1/2n then from 1/(2n +1) to 1/3n then 1/(3n+1) to 1/4n.

Each of these sections has n terms in it. So sum of first section is. > n/2n, second section sum >n/3n , third section sum greater than n/4n. When you add these three up you get sum greater than 1/2 +1/3 +1/4 = 13/12

Answer 3

A monotonic sequence either continually increases or continually decreases.

Take the rth term of your sequence 1/(n + r) and (r + 1)th term 1/(n + r + 1)

now 1/(n + r + 1) - 1/(n + r) = [(n+r) - (n+r+1)]/(n + r)(n+r+1)]

= -1/[(n + r)(n + r + 1)] < 0 for all n,r so the sequence is continually decreasing.

Limits

our sequence has a last term of 1/(n + 3n) so there are 3n terms in the sequence.
Your sequence is at all times less than 1/n + 1/n + 1/n ....+1/n
This sequence has a limit of 3n x 1/n = 3

Your L is <3

Your sequence is at all times greater than

1/(4n) + 1/(4n) + ...+ 1/(4n) 3n terms so your L > 3n/4n

So I make it 3/4 < L < 3

I haven't yet found a closer sequence...but I hope you get the idea

Answer 4

L