How prove this limit

Solution 1:

$\def\floor#1{\lfloor #1\rfloor} \def\ceil#1{\lceil #1\rceil}$ This is really a question about manipulating sums, floors, and ceilings, and one book that discusses these things well is "Concrete Mathematics".

Consider the first sum $S$: $$ S = \sum_{0\leq k\leq \lfloor n\phi\rfloor} \frac{1}{\lfloor k/\phi\rfloor+2}, $$ and eliminate the floor function by introducing a new variable $j$, such that $j=\floor{k/\phi}$ whenever $j\leq k/\phi<j+1$ (by definition of floor): $$ \sum_{0\leq k\leq \floor{n\phi}} \sum_j \frac{1}{j+2}[\phi j\leq k<(j+1)\phi]. $$ The condition in the Iverson bracket there is equivalent to $$[\ceil{j\phi}\leq k<\ceil{(j+1)\phi}]. $$ This is a range of $k$, of length $\ceil{(j+1)\phi} - \ceil{j\phi}$, and these intervals, for all $0\leq j<n$ fit completely within $0\leq k\leq \floor{n\phi}$ and cover it totally. So the sum is equal to $$ \sum_{0\leq j<n} \frac{\ceil{(j+1)\phi} - \ceil{j\phi}}{j+2}. $$ This sum telescopes a litle, so we can shift $j$ down by one in the first term to get $$ \frac{\ceil{n\phi}}{n+1} + \sum_{1\leq j<n} \frac{\ceil{j\phi}}{(j+1)(j+2)}. $$

Finally, the entire limit is $$ \frac{\ceil{n\phi}}{n+1} + \sum_{1\leq j<n} \frac{\ceil{j\phi}-\floor{j\phi}}{(j+1)(j+2)}. $$ Because $\phi$ is irrational, the numerator of the summands is always $1$, and $$ \sum_{1\leq k<n} \frac{1}{(k+1)(k+2)} = \frac12 - \frac1{1+n}. $$ Hence the expression in the limit is $$ \frac{\ceil{n\phi}}{n+1} + \frac12 + O(n^{-1}) = \phi+\frac12 + O(1/n), $$ so the limit is $$ \frac12+\phi. $$