Lower Bound

Vassil Vassilev , a 14 year old Bulgarian student from St Michael's College, Leeds, sent the following solution:

Element No 1

$${1\over 2} + {2\over 1} = {1 + 4\over 2} = {5\over 2} = 2 + {1\over 2} = 2 + {1\over 1\times(1+1)}$$ Element No 2 $${2\over 3} + {3\over 2} = {4 + 9\over 6} = {13\over 6} = 2 + {1\over 6} = 2 + {1\over 2\times(2+1)}$$ Element No 3 $${3\over 4} + {4\over 3} = {9 + 16\over 12} = {25\over 12} = 2 + {1\over 12} = 2 + {1\over 3\times(3 + 1)}$$ Element No 4 $${4\over 5} + {5\over 4} = {16 + 15\over 20} = {41\over 20} = 2 + {1\over 20} = 2 + {1\over 4\times(4 + 1)}$$ Element No n

Vassil stopped here but do you notice that, as $n$ gets bigger and bigger, so $${1\over n(n + 1)}$$ gets smaller and smaller and closer and closer to zero? If this sequence went on for ever the terms would get closer and closer to 2 without ever actually being equal to 2. Another way of seeing this is to look at $${n\over n + 1}$$ and see that this must get closer and closer to the value 1 as $n$ gets bigger and bigger and also $${n + 1\over n}$$ must also get closer and closer to 1, so the $n$th term of the sequence, namely $${n\over n + 1} + {n + 1\over n}$$ must get closer and closer to 2. We say the limit of this sequence, as $n$ tends to infinity, is 2.

Some, but not all, of the points on the graph of

$$y = x + {1\over x}$$ represent terms of the sequence. You might like to draw the graph and look at what happens to it around $x = 1$.

Vassil worked out the $n$th term of the similar sequence formed by adding the squares of these fractions and after doing some algebra to simplify the expression he obtained the following result:

$$\left({n\over n + 1}\right)^2 + \left({n + 1\over n}\right)^2 = 2 + \left({2n + 1\over n(n + 1)}\right)^2.$$ Here again the limit as n tends to infinity is 2. You might like to prove this result in another way.