Lower bound

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

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

Vassil stopped here but do you notice that, as $n$ gets bigger and bigger, so $$\frac{1}{n(n+1)}$$ 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 $$\frac{n}{n+1}$$ and see that this must get closer and closer to the value 1 as $n$ gets bigger and bigger and also $$\frac{n+1}{n}$$ must also get closer and closer to 1, so the $n$th term of the sequence, namely $$\frac{n}{n+1}+\frac{n+1}{n}$$ 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 $$y=x+\frac{1}{x}$$ 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: $${\left(\frac{n}{n+1}\right)}^2+{\left(\frac{n+1}{n}\right)}^2=2+{\left(\frac{2n+1}{n(n+1)}\right)}^2.$$ Here again the limit as n tends to infinity is 2. You might like to prove this result in another way.