# Convergence of $S_n(a) = \sum_{an < k \leqslant (a+1)n} \frac{1}{\sqrt{kn-an^2}}$

Let $a\in\mathbb{R}$ et $n \in\mathbb{N}$,

Denote the following sequence, $$\displaystyle S_n(a) = \sum_{an < k \leqslant (a+1)n} \frac{1}{\sqrt{kn-an^2}}$$

For which values ​​of $a$ the sequence $(S_n(a))$ converges?

With an Integral test we can prove that $$\sum_{an+1 < k \leq an+n} \frac{1}{\sqrt{kn-an^2}} \to 2$$

However, I don't see how can I continue ?

EDIT: I forgot to say $k$ is an integer. My apology.

I assume that $k$ shall be an integer, otherwise it would be unclear which set it should traverse.

With $j = k - \lfloor an\rfloor$, we have

$$S_n(a) = \sum_{an < k \leqslant (a+1)n} \frac{1}{\sqrt{kn-an^2}} = \sum_{j=1}^n \frac{1}{\sqrt{(\lfloor an\rfloor + j)n - an^2}} = \sum_{j=1}^n \frac{1}{\sqrt{(j-(an-\lfloor an\rfloor))n}}.$$

Let us write $\rho_n(a) = an - \lfloor an\rfloor$. By definition, we have $0 \leqslant \rho_n(a) < 1$.

For the sum without the first term, we have the easy estimates

$$\sum_{j=2}^n \frac{1}{\sqrt{jn}} \leqslant \tilde{S}_n(a) := \sum_{j=2}^n \frac{1}{\sqrt{(j-\rho_n(a))n}} < \sum_{j=2}^n \frac{1}{\sqrt{(j-1)n}}$$

where both outer terms tend to $\int_0^1\frac{dx}{\sqrt{x}} = 2$.

So the convergence of $S_n(a)$ is entirely determined by the behaviour of the first term

$$\frac{1}{\sqrt{(1-\rho_n(a))n}}.$$

If there is a $c > 0$ such that $1-\rho_n(a) \geqslant c$ for all $n$, then the first term is at most $\frac{1}{\sqrt{cn}}$ and tends to $0$, whence in that case $S_n(a) \to 2$. For rational $a$ such a $c > 0$ exists, so $S_n(a)$ converges (to $2$) for all rational $a$.

But for irrational $a$, if we denote the convergents of the continued fraction of $a$ by $\frac{p_m}{q_m}$, we have

$$\left\lvert a - \frac{p_m}{q_m}\right\rvert < \frac{1}{q_m^2}.$$

The convergents with odd indices are all $> a$, thus

$$0 < \frac{p_{2m+1}}{q_{2m+1}} - a < \frac{1}{q_{2m+1}^2} \iff 0 < p_{2m+1} - q_{2m+1}a < \frac{1}{q_{2m+1}},$$

and we see that $\lfloor q_{2m+1} a\rfloor = p_{2m+1}-1$ and

$$1 - \rho_{q_{2m+1}}(a) = 1+\lfloor q_{2m+1}a\rfloor - q_{2m+1}a = p_{2m+1}- q_{2m+1}a,$$

and so

$$0 < (1-\rho_{q_{2m+1}}(a))q_{2m+1} < 1,$$

which means that the first term in $S_n(a)$ is greater than $1$ infinitely often. But for the convergents with even indices, we have

$$0 < \rho_{q_{2m}}(a) = q_{2m}a - p_{2m} < \frac{1}{q_{2m}},$$

and thus $1-\rho_n(a) > \frac{1}{2}$ for infinitely many $n$, so for irrational $a$, we have

$$2 = \liminf_{n\to\infty} S_n(a) < 3 \leqslant \limsup_{n\to\infty} S_n(a).$$

Result:

$S_n(a)$ converges if and only if $a \in\mathbb{Q}$.

• Very very nice! I am confused that not seeing @Lucian assumed that $k$ is a real. Thank you very much!
– user117932
Commented Apr 6, 2014 at 13:08

$k=an+j$, with $j\in\overline{1\ldots n}=>S_n(a)=\displaystyle\sum_{j=1}^n\dfrac1{\sqrt{jn}}=\dfrac1n\cdot\sum_{j=1}^n\dfrac1{\sqrt{\dfrac jn}}$ , which tends to $\displaystyle\int_0^1\frac{dx}{\sqrt x}$ as n tends to infinity, whose value is $2$. See Riemann sum for more details.

• Why not ? a and k may be real, but n and j are always positive integers. Commented Apr 5, 2014 at 13:04
• There's nothing wrong with the substitution, nor does $S(\sqrt2)$ diverge. Commented Apr 5, 2014 at 13:37
• I meant that your observation is simply incorrect. Obviously, $1<\sqrt2<2$, so, by the squeeze theorem, $S(\sqrt2)$ lies in between $S(1)$ and $S(2)$. Since the latter two are convergent, then so is the former. Commented Apr 5, 2014 at 13:48
• No, $S(a)$ doesn't "obviously" depend on a, for the same reason that a (constant) function does not "obviously" or necessarily depend on its argument. Just write out the first few terms of the sum, and see it for yourself. As for the squeezing part, since $1<\sqrt2<2$ it is quite trivial to show that $S(\sqrt2)$ lies in between $S(1)$ and $S(2)$. Commented Apr 5, 2014 at 14:38
• @Lucian Though it is not explicitly said anywhere, I'm rather convinced that $k$ shall be an integer. Thus $k$ would take the values $\lfloor an\rfloor + 1\, \lfloor an\rfloor + 2,\dotsc, \lfloor an\rfloor + n$. It would be unusual to use $k$ for it otherwise. Commented Apr 6, 2014 at 13:00