Asymptotic bound for a particular sum

In a post on his blog, Qiaochu Yuan asks how fast the sequence \[ R_n = \frac{n + \sqrt{n} + \sqrt[3]{n} + \cdots + \sqrt[n]{n}}{n}. \] converges to its limit. If you want to calculate this limit then you should stop reading here. Anyhow, the limit is two, so if the asymptotic behavior of \[ S_n = R_n - 2 + \frac{1}{n} = \frac{1}{n} \sum_{k = 2}^{n} \left(n^{1/k} - 1\right) \] is any slower than $O(1/n)$, then that behavior must belong to the quantity $R_n - 2$.

In Qiaochu's post he gives evidence for why he believes the asymptotic behavior to be $O(1/\sqrt{n})$. We'll show that this is indeed the case. To do this we'll need a small lemma.

- (a) $\lim_{x \to 0^+} \phi(x) = \infty$,
- (b) $\lim_{x \to \infty} \phi(x) = 3/2$,
- (c) There exists an $x_0 \in (0,\infty)$ such that $\phi$ is decreasing on $(0,x_0)$ and increasing on $(x_0,\infty)$ and $\phi(x_0) < 3/2$.

A plot of $\phi(x)$ (in blue) and its horizontal asymptote $3/2$
(dashed and red).

**Proof.** (a) This follows from the
application of
L'Hopital's Rule.

(b) We have
\[
\begin{align*}
\phi(1/x) &= \frac{e^x - 1}{x^2} - \frac{1}{x(x+1)} \\
&= \frac{x + \frac{x^2}{2} + O\left(x^3\right)}{x^2} -
\frac{1}{x}\left(1 - x + O\left(x^2\right)\right) \\
&= \frac{3}{2} + O(x) \end{align*}
\]
as $x \to 0$, so
\[
\phi(x) = \frac{3}{2} + O\left(\frac{1}{x}\right)
\]
as $x \to \infty$.

(c) We will show that
\[
\phi'(x) = \frac{1}{(x+1)^2}\left(e^{1/x} (x + 1)^2 (2x - 1) - x
\left(2x^2 + 5x + 4\right)\right)
\]
has exactly one positive real zero. Hence we consider the function
\[
f(x) = (x+1)^2 \phi'(x) = e^{1/x} (x + 1)^2 (2x - 1) - x \left(2x^2 +
5x + 4\right).
\]
First, $f(1) < 0 < f(2)$, so $f(x)$ has a zero in the
interval $(1,2)$ with the desired sign change. The result will follow
if
\[
\tag{1}
f'(x) = (x + 1) \left(e^{1/x} \left(6x - 2 - \frac{1}{x} +
\frac{1}{x^2}\right) - 6x - 4\right) > 0
\]
when $x > 0$. Well, if we define
\[
g(x) = \frac{x^2 f'(1/x)}{x+1} = e^x \left(x^3 - x^2 - 2x + 6\right) -
4x - 6
\]
then $g(0) = 0$, so $(1)$ will follow if
\[
\tag{2}
g'(x) = e^x \left(x^3 + 2x^2 - 4x + 4\right)-4 > 0
\]
when $x > 0$. Since $g'(0) = 0$, we repeat the process once more
to calculate
\[
g''(x) = e^x x^2 (x + 5).
\]
Thus $g''(x) > 0$ when $x > 0$, giving inequality
$(2)$ which in turn implies inequality $(1)$.

In light of this conclusion, the facts $\phi(1) < 3/2$
and $\phi'(1) < 0$ imply that $\phi(x_0) < 3/2$.

The lemma tells us all we will need to know about $\phi$. In particular, we can conclude that, if $\phi(r) < 3/2$ for some $r$, then $\phi(x) < 3/2$ for all $x \geq r$. Therefore, we can take $r$ small enough so that $\phi(x) \leq \phi(r)$ for $x \geq r$. We can now derive the desired bound.

**Proof.** By the lemma, $\phi(x) \leq \phi(2/\log n)$
for $x \geq 2/\log n$ when $n$ is large enough. Applying the map $x
\mapsto x/\log n$ renders the inequality
\[
n^{1/x} - 1 \leq \log n \left(\frac{\phi(2/\log n) \log n}{x^2} +
\frac{1}{x + \log n}\right)
\]
for $x \geq 2$.

The summand $n^{1/k} - 1$ is a decreasing function of $k$, so we
canwrite
\[
\begin{align*}
S_n &\leq \frac{1}{n} \left(\sqrt{n} - 1 + \int_2^n
\left(n^{1/x} - 1\right) dx \right) \\
&\leq \frac{1}{n} \left(\sqrt{n} - 1 + \int_2^n \log n
\left(\frac{\phi(2/\log n) \log n}{x^2} + \frac{1}{x + \log n}\right)
dx \right) \\
&= \frac{1}{n} \left(\sqrt{n} - 1 + \frac{\phi(2/\log
n)(n-2)(\log n)^2}{2n} + \log n \log\left(\frac{n + \log n}{2 + \log
n}\right) \right).
\end{align*}
\]
(Justification for the first inequality can be found here.)
Plugging
in the value of $\phi(2/\log n)$ we get
\[
\begin{split}
S_n &\leq \frac{8 - 8\sqrt{n} - 6n + 6n^{3/2} + 8\log n -
4\sqrt{n}\log n - 5n\log n + 3n^{3/2}\log n}{n^2 (2 + \log n)} \\
&\hspace{3in} + \frac{\log n \log\left(\frac{n + \log
n}{2 + \log n}\right)}{n}.
\end{split}
\]
The first term is $O \left(\frac{1}{\sqrt{n}}\right)$, and, since
\[
\frac{\log n \log\left(\frac{n + \log n}{2 + \log
n}\right)}{\sqrt{n}} \to 0,
\]
so is the second.

Some comments on approximating $e^{1/x}-1$

The lemma tells us that, for $x$ large enough (e.g. for $x \geq 103/223$), \[ \phi(x) < 3/2, \] and so \[ \tag{3} e^{1/x} - 1 < \frac{1}{x+1} + \frac{3}{2x^2}. \] Hence the coefficient on $1/x^2$ is significant geometrically; it is the upper bound for $\phi$ on the interval in question. Interestingly, it also plays an important analytical role. To see this, let's take a step back and try "guessing" the above relationship.

We want to approximate the function \[ e^{1/x} - 1 = \frac{1}{x} + \frac{1}{2x^2} + \frac{1}{6x^3} + \cdots \] for $x$ large enough. And not only this; our purposes require that the function we approximate it by have a known integral. Further, we would like it to have the same behavior near infinity, so we want it to be equal up to at least inverse-first order. If we apply the map $x \mapsto 1/x$, our question becomes that of approximating \[ e^x - 1 = x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots \] near the origin by some function equal to at least first order and with a known integral (after transforming back).

Because every coefficient in the power series of the exponential function is positive, we will always have \[ \sum_{k=1}^{r} \frac{x^k}{k!} < e^x - 1, \] so we'll need to look elsewhere if we want to bound our function above. The function \[ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots \] seems like a good candidate: we can simply multiply by $x$ to get equality to first order and it certainly has a known integral after the appropriate transformation. However, things blow up when $x$ is near unity, and has the wrong sign when $x$ is larger. These clash with properties of the exponential function we'd like to preserve. Let's try \[ \frac{1}{x+1} = 1 - x + x^2 - x^3 + \cdots. \] Multiplying this by $x$ we do indeed get equality up to first order near zero, and its behavior elsewhere doesn't conflict nearly as badly as the previous guess. Transforming back for a moment, we have \[ e^{1/x} - 1 - \frac{1}{x+1} = \frac{c}{x^2} + O\left(\frac{1}{x^3}\right) \] for some constant $c$, so that \[ x^2\left(e^{1/x} - 1 - \frac{1}{x+1}\right) \to c \] as $x \to \infty$. The expression on the left is exactly $\phi(x)$, so $c = 3/2$. Thus this choice of coefficient for $1/x^2$ in $(3)$ ensures equality up to inverse-second order near infinity. In general, if we compute the difference \[ \begin{align*} e^x - 1 - \frac{x}{x+1} &= \sum_{k = 2}^{\infty} \left(\frac{1}{k!} + (-1)^k\right) x^k \\ &= \frac{3}{2}x^2 - \frac{5}{6}x^3 + \frac{25}{24}x^4 + \cdots, \end{align*} \] we may take as many terms as we like to get a better approximation of the exponential function. However, it seems that we must take up to and including an even power in order to get the correct inequality.

Antonio R. Vargas

July 25, 2011

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