I accidentally observed that $\sqrt{n} - \sqrt{n-1}$ tends to $0$ for higher values of $n$, so I've decided to try to sum it all, but that sum diverged. So I've tried to make it converge by giving it a power $m$.

$$S_m=\sum_{n=1}^\infty (\sqrt{n} - \sqrt{n-1})^m$$

How would one calculate the values of this sum for a choosen $m\in\mathbb R$?

Not just estimate but write them in a non-decimal form if possible, preferably using a better converging formula.

It converges if $m>2$.

The values seem to tend to numbers with non-repeating decimals.

  • Thanks to achille hui and his partial answer here, it looks like $S_m$ for odd values of $m$ is a linear combination of riemann zeta function at negative half-integer values:

\begin{align} S_3 &\stackrel{}{=} -6\zeta(-\frac12) \approx 1.247317349864128...\\ S_5 &\stackrel{}{=} -40\zeta(-\frac32) \approx 1.019408075593322...\\ S_7 &\stackrel{}{=} -224\zeta(-\frac52) - 14\zeta(-\frac12) \approx 1.00261510344449... \end{align}

If we decide to replace the constant $m$ with $n\times{k}$ where $k$ is a new constant, then we can talk about $S_k$, Which converges if $k>0$;

$$S_k=\sum_{n=1}^\infty (\sqrt{n} - \sqrt{n-1})^{nk}$$

I wonder if these values could also be expressed in a similar way as $S_m$.

Values still tend to numbers with seemingly non-repeating decimals according to the Wolfram Alpha:

$$ S_1 \approx 1.20967597004937847717395464774494290 $$

Also notice that the functions of these sums are similar to the zeta function; $\color{blue}{S_m} \sim \color{red}{\zeta}$

enter image description here

But I think it's only due the fact that they all approach $1$?

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    $\begingroup$ When you choose $m = n$, your power is no longer a constant and is changing constantly, which is a different problem then choosing $m$ as some number. $\endgroup$ – Mitch Dec 2 '16 at 20:33
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    $\begingroup$ By manipulating the sums as divergence series. I obtain a bunch of expressions for odd $m$. They do seem to match the actual sums numerically. $$\begin{align} S(3) &\stackrel{?}{=} -6\zeta(-\frac12) \approx 1.247317349864128,\\ S(5) &\stackrel{?}{=} -40\zeta(-\frac32) \approx 1.019408075593322,\\ S(7) &\stackrel{?}{=} -224\zeta(-\frac52) - 14\zeta(-\frac12) \approx 1.00261510344449 \end{align} $$ $\endgroup$ – achille hui Dec 2 '16 at 20:53
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    $\begingroup$ @SimpleArt I ask WA to compute the sum for $m = 3$ with up to $10^{15}$ terms. It matches the number I quote up to $8$ decimals places... $\endgroup$ – achille hui Dec 2 '16 at 21:39
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    $\begingroup$ @tired, using the asymptotic expansion of sum of square roots in this answer, $$\sum_{k=1}^n k^{1/2} \asymp \frac23 n^{3/2} + \frac12 n^{1/2} + \zeta(-1/2) + \cdots$$ One can show $S(3)$ is indeed $-6\zeta(-1/2)$. I suspect there is a similar asymptotic expansion $$\sum_{k=1}^n k^{3/2} \asymp \frac25 n^{5/2} + \frac12 n^{3/2} + \frac18 n^{1/2} + \zeta(-3/2)$$ I don't know how to prove this but if this is true, $S(5)$ also equals to $-40\zeta(-3/2)$. It is too bad I don't know Merlin transform enough to justify above expansion. $\endgroup$ – achille hui Dec 3 '16 at 1:17
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    $\begingroup$ @tired, I find the general version of asymptotic formula one need $$\sum_{k=1}^n k^\alpha - \zeta(-a) \asymp \frac{n^{\alpha+1}}{\alpha+1} \sum_{s=0}^\infty \binom{\alpha+1}{s}\frac{(-1)^sB_s}{n^s} $$ in an exercise of Olver's book "Asymptotics and Special functions". I've checked, for all odd $m \le 13$, $S(m)$ can be expressed in terms of zeta functions at negative half-integer values. $\endgroup$ – achille hui Dec 3 '16 at 17:53

This is a partial answer. It attempt to justify some expressions I obtained for odd $m$ by manipulating the sums as divergence series.

What I have shown so far is for all odd $m \ge 3$ and $\le 99$, following sums

$$S_m \stackrel{def}{=} \sum_{n=1}^\infty (\sqrt{n} - \sqrt{n-1})^m$$

is a linear combination of zeta function at negative half-integer values.

For any $m > 0$, we have

$$(\sqrt{n} - \sqrt{n-1})^m = T_m(\sqrt{n}) - U_{m-1}(\sqrt{n})\sqrt{n-1}$$

where $T_m(x)$, $U_{m-1}(x)$ are the Chebyshev polynomials of first and second kind.

When $m = 2\ell+1$ is odd, $T_m(x)$ is an odd polynomial in $x$ with degreee $2\ell+1$ and $U_{m-1}(x)$ is an even polynomial in $x$ with degree $2\ell$. We can rewrite the partial sum of $S_m$ as

$$\begin{align} S_{m,p} \stackrel{def}{=} \sum_{n=1}^p (\sqrt{n}-\sqrt{n-1})^m & = \sum_{n=1}^p \left(T_m(\sqrt{n})-U_{m-1}(\sqrt{n})\sqrt{n-1} \right)\\ & = \left(\sum_{n=1}^p \left(T_m(\sqrt{n})-U_{m-1}(\sqrt{n+1})\sqrt{n}\right)\right) + U_{m-1}(\sqrt{p+1})\sqrt{p} \end{align} $$ What's inside the sum of last line will be $\sqrt{n}$ times a polynomial in $n$ of degree $\ell$.
Let $\alpha_0, \alpha_1, \cdots \alpha_\ell$ be the coefficients of this polynomial. i.e. $$T_m(\sqrt{n})-U_{m-1}(\sqrt{n+1})\sqrt{n} = \sum_{k=0}^\ell \alpha_k n^{k+1/2}$$ In terms of them, we have $$S_{m,p} = \sum_{k=0}^\ell \alpha_k \sum_{n=1}^p n^{k+1/2} + U_{2\ell}(\sqrt{p+1})\sqrt{p}$$

For any $s > 0, \notin \mathbb{Z}$, we have following asymptotic expansion${}^{\color{blue}{[1]}}$

$$\sum_{n=1}^p n^s - \zeta(-s) \asymp \frac{1}{s+1}\sum_{k=0}^\infty \binom{s+1}{k} (-1)^k B_k p^{s+1-k}$$

Let $\Lambda(s,p)$ be the finite sum of those terms on RHS which diverges as $p \to \infty$.

$$\Lambda(s,p) \stackrel{def}{=} \frac{1}{s+1}\sum_{k=0}^{\left\lfloor s + 1 \right\rfloor} \binom{s+1}{k} (-1)^k B_k p^{s+1-k}$$

By construction, these counter-terms will kill the divergence in the sum $\sum\limits_{n=1}^p n^s$ as $p \to \infty$.
More precisely, we have

$$\lim_{p\to\infty} \left( \sum_{n=1}^p n^s - \Lambda(s,p) \right) = \zeta(-s)$$

Rewrite the partial sums again, we have

$$S_{m,p} = \sum_{k=0}^\ell \alpha_k \left( \sum_{n=1}^p n^{k+1/2} - \Lambda(k+1/2,p)\right) + \underbrace{ \sum_{k=0}^\ell \alpha_k \Lambda(k+1/2,p) + U_{2\ell}(\sqrt{p+1})\sqrt{p}}_{R(m,p)}$$

If the piece $R(m,p)$ in above expression vanishes identically, we will have

$$S_m = \lim_{p\to\infty} S_{m,p} = \sum_{k=0}^\ell \alpha_k \zeta(-(k+1/2))$$

I don't know how to prove $R(m,p)$ vanish for general $m$. However, for odd $m \le 99$, I have used a CAS to compute $R(m,p)$ symbolically and verify all of them vanish identically.

As a result, for odd $m \le 100$, $S_m$ is a linear combination for $\zeta(z)$ at negative half-integer values.
For reference, following is a short list of $S(m)$ for small odd $m$.

$$ \begin{align} S_3 &= -6 \zeta(-1/2)\\ S_5 &= -40 \zeta(-3/2)\\ S_7 &= -14 \zeta(-1/2) -224 \zeta(-5/2) \\ S_9 &= -240 \zeta(-3/2) -1152 \zeta(-7/2)\\ S_{11} &= -22 \zeta(-1/2) -2464 \zeta(-5/2) -5632 \zeta(-9/2)\\ S_{13} &= -728 \zeta(-3/2) -19968 \zeta(-7/2) -26624 \zeta(-11/2)\\ S_{15} &= -30 \zeta(-1/2) -12096 \zeta(-5/2) -140800 \zeta(-9/2) -122880 \zeta(-13/2)\\ \end{align} $$


  • $\color{blue}{[1]}$ - I found this expansion in an exercise of Frank W.J. Olver's book "Asymptotics and Specical Functions". Look at $\S 8.3$ "Contour integral for the remainder term" for more info on this type of expansion.
  • $\begingroup$ Achille Hui, how can i reach you? any contacts? I need help on a previous question of mine that you answered. $\endgroup$ – Iconoclast Dec 4 '16 at 11:06
  • $\begingroup$ I can't fully yet grasp your expressions, so may I ask how could I simply generate the factors for zeta functions? ($a_k$ if I'm not mistaken) $\endgroup$ – Vepir Dec 9 '16 at 8:11
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    $\begingroup$ @Matta If you expand $$ \begin{align} & \frac{1}{\sqrt{n}}\left[T_m(\sqrt{n}) - U_{m-1}(\sqrt{n+1})\sqrt{n}\right]\\ = & \frac{1}{2\sqrt{n}}\left[ (\sqrt{n}-\sqrt{n-1})^m + (\sqrt{n}+\sqrt{n-1})^m + (\sqrt{n+1} - \sqrt{n})^m - (\sqrt{n+1} + \sqrt{n})^m \right] \end{align}$$ into a polynomial in $n$, $\sum\limits_{k=0}^\ell \alpha_k n^k$. The coefficients $\alpha_k$ will be the factor in front of $\zeta(-k-1/2)$ $\endgroup$ – achille hui Dec 9 '16 at 8:37

It does converge for $m > 2$, since $\sqrt{n} - \sqrt{n-1} \sim 1/(2\sqrt{n})$ and $\sum_n n^{-m/2}$ converges for $m > 2$.


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