Increasing or decreasing order

Number Theory Level 2

The nth element of a set R is equal to n/n+1 where n = 0,1,2,3,4,..... The first few elements of R are (0,1/2,2/3,3/4...) What can be said about the elements of the set R ?.

The order of elements is always decreasing. They are in increasing order from the first to the last element The order of elements from smallest to greatest is random, Some elements of R cannot be calculated.

Srinivasa Gopal by Srinivasa Gopal , 53, India

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3 solutions

Mr. India
Mar 17, 2019

T n = n n + 1 T_n=\frac{n}{n+1}

T n + 1 = n + 1 n + 2 T_{n+1}=\frac{n+1}{n+2}

T n + 1 − T n = n + 1 n + 2 − n n + 1 T_{n+1}-T_n=\frac{n+1}{n+2}-\frac{n}{n+1}

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

= n 2 + 2 n + 1 − n 2 − 2 n n 2 + 3 n + 2 =\frac{n^2+2n+1-n^2-2n}{n^2+3n+2}

1 n 2 + 3 n + 2 > 0 \frac{1}{n^2+3n+2}>0

So, T n + 1 > T n T_{n+1}>T_n

And sequence is always i n c r e a s i n g \boxed{\boxed{increasing}}

1 Helpful 0 Interesting 1 Brilliant 0 Confused
Jordan Cahn
Mar 18, 2019

1 − n n + 1 = 1 n + 1 > 1 n + 2 = 1 − n + 1 n + 2 1-\frac{n}{n+1} = \frac{1}{n+1} > \frac{1}{n+2} = 1-\frac{n+1}{n+2}

Therefore n n + 1 < n + 1 n + 2 \dfrac{n}{n+1}<\dfrac{n+1}{n+2} and the sequence is increasing .

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Srinivasa Gopal
Mar 17, 2019

The successive elements of set R can be written as n/n+1 and n+1/n+2 . Let us calculate(( n+1)/(n+2) - n/(n+1)) , the difference works out to be 1/(n+1)(n+2) which is always greater than 0. So the order of elements of R from n = 0 to n = inf is always increasing,

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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