Real Analysis
297.Examining the sequence, we see that themth term of the sequence is equal ton
exactly for thosemthat satisfy
n^2 −n
2+ 1 ≤m≤n^2 +n
2.
So the sequence grows about as fast as the square root of twice the index. Let us rewrite
the inequality as
n^2 −n+ 2 ≤ 2 m≤n^2 +n,then try to solve forn. We can almost take the square root. And becausemandnare
integers, the inequality is equivalent to
n^2 −n+1
4
< 2 m<n^2 +n+1
4
.
Here it was important thatn^2 −nis even. And now wecantake the square root. We
obtain
n−1
2
<
√
2 m<n+1
2
,
or
n<√
2 m+1
2
<n+ 1.Now this happens if and only ifn=
√
2 m+^12 , which then gives the formula for the
general term of the sequence
am=⌊√
2 m+1
2
⌋
,m≥ 1.(R. Graham, D. Knuth, O. Patashnik,Concrete Mathematics: A Foundation for
Computer Science, 2nd ed., Addison–Wesley, 1994)