# Convergence of $u_n$ defined by $\sum\limits_{k=1}^{u_n-1}\frac{1}k\leq n<\sum\limits_{k=1}^{u_n}\frac{1}k$

For every positive integer $n$, define the positive integer $u_n$ by the condition that $$\sum\limits_{k=1}^{u_n-1}\frac{1}k\leq n<\sum\limits_{k=1}^{u_n}\frac{1}k$$ Let $w_n=\frac1{u_n}$. Does the sequence $(w_n)$ converge and how to prove rigorously it does?

What I did:

I proved that $u_n$ exists and is unique, and that $u_1=2$ and $u_2=4$.

Proof of uniqueness:

Suppose that $u_n>v_n$ satisfying the defining condition, then $v_n\leq u_n-1$ so $\sum\limits_{k=1}^{v_n}\frac{1}k\leq n$, which is absurd, thus $u_n$ = $v_n$.

• $u_n\to+\infty$, certainly! Jun 12 '14 at 15:19
• Did you mean convergence of $u_n$ (small lettered)? Jun 12 '14 at 15:19
• @PedroTamaroff; @boywholived I made an error, let me correct that Jun 12 '14 at 15:20
• After Pedro Tamaroff's hint that $u_n\rightarrow +\infty$, observe that $\dfrac{1}{u_n}\cdot u_n =1 \rightarrow 1$, then you can use this to prove that $\dfrac{1}{u_n}$ converges to $0$. Jun 12 '14 at 15:28
• Actually $u_1$ is not unique and to prove that $u_n$ is unique for every $n\geqslant2$ requires to prove that $\sum\limits_{k=1}^K\frac1k$ is not an integer when $K\geqslant2$ (did you really prove this?).
– Did
Jun 12 '14 at 15:32

$$n\lt\sum_{k=1}^{u_n}\frac1k\leqslant1+\int_1^{u_n}\frac{\mathrm dt}t=1+\log u_n\implies\frac1{u_n}\lt\frac{\mathrm e}{\mathrm e^n}$$