Challenge Level

**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

\begin{eqnarray} {n\over n + 1} + {n+1\over n}
&=& {n^2 + (n + 1)^2\over n(n + 1)} = {n^2 + n^2 + 2n +
1\over n^2 + n}\\ &=& {2n^2 + 2n\over n^2 + n} + {1\over
n^2 + n} = 2 + {1\over n^2 + n}\\ &=& 2 + {1\over n(n + 1)}
\end{eqnarray}

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.