Does there exist a closed form for $L_k$ for any $k>3$?

Question

I defined a sequence $L_k$ as the limit of a sequence of "hyperharmonic" series in this question.

I was surprised to find that $L_3=(\sqrt{13+4\sqrt2}-1)/2$, but was unable to find a representation for any $k>3$.

Does there exist a closed form for $L_k$ for any $k>3$?

Definition:

Let $H_k^{-1}=1$ for every positive integer $k$, and

$$H_k^r=\sum_{n=1}^k\left(\sum_{m=1}^nH_m^{r-1}\right)^{-1}$$

for every positive integer $k$ and non-negative integer $r$, so that $$H_k^0=\sum_{n=1}^k\dfrac1n$$ coincides with the usual definition of the $k$th harmonic number. I defined

$$L_k=\lim_{r\to\infty}H_k^r$$

Clearly $L_1=1$, it can be deduced from the continued fraction for $\sqrt2$ and the above definitions that $L_2=\sqrt2$, and I found the above value for $L_3$ with the use of high precision floating point.

Edit: I guess it's obvious that $L_k$ satisfies

$$L_k=\sum_{n=1}^k\left(\sum_{m=1}^nL_m\right)^{-1}$$

but I don't know how to apply this to calculating $L_k$. Wolfram Alpha was able to give me a hideous expression for

$L_4={\small-\dfrac{5+3\sqrt2+\sqrt{13+4\sqrt2}+\sqrt{26+8\sqrt2}-\sqrt2(361+178\sqrt2+51 \sqrt{13+4\sqrt2}+64\sqrt{26+8\sqrt2})}{2(1+2\sqrt2+\sqrt{13+4\sqrt2})}}$

Edit: I'm not so sure of this last sum. I believe I was given a correct result by Wolfram Alpha, but I think I might've made an error in transcription. Unfortunately I'm having a hard time getting the result again.

I put the simpler formula $L_k=\dfrac1{L_k-L_{k-1}}-\dfrac1{L_{k-1}-L_{k-2}}$ into Wolfram Alpha and it gave me the recurrence

$${\scriptsize L_k=\dfrac{L_{k-1}^2-L_{k-1}L_{k-2}+\sqrt{L_{k-1}^4-2L_{k-1}^3L_{k-2}+L_{k-1}^2L_{k-2}^2+6L_{k-1}^2-10L_{k-1}L_{k-2}+4L_{k-2}^2+1}-1}{2(L_{k-1}-L_{k-2})}}$$

which seems to work.

Answer 1

Consider $U_n := \sum_{k\leq n}L_k$ which is a simpler object. It forms an increasing sequence, and from your recurrence relation, $$ U_n - U_{n-1} = \sum_{j\leq n} 1/U_j.$$ Assume that $(L_k)$ is bounded, say by $C>0$. Then $U_n \leq Cn$, and $$ L_k = \sum_{j\leq k} 1/U_j \geq (\log k)/C $$ which is a contradiction as $k\to\infty$. You can probably tweak this to an asymptotic formula for $L_k$.

Answer 2

Here is some additional information which could be helpful.

Starting with OPs definition of $H_k^r$: \begin{align*} &H_k^{-1} := 1\qquad& k\geq 1\tag{1}\\ &H_k^r:=\sum_{n=1}^k\frac{1}{H_1^{r-1}+\ldots+H_n^{r-1}}\qquad &k\geq 1, r\geq 0\tag{2}\\ \text{and}\quad&\\ &L_k:=\lim_{r\rightarrow\infty}H_k^r\qquad&k\geq 1\tag{3} \end{align*} We observe according to (2) \begin{align*} H_{k+1}^r&=\sum_{n=1}^{k+1}\frac{1}{H_1^{r-1}+\ldots+H_n^{r-1}}\\ &=\left(\sum_{n=1}^{k}\frac{1}{H_1^{r-1}+\ldots+H_n^{r-1}}\right)+\frac{1}{H_1^{r-1}+\ldots+H_{k+1}^{r-1}}\qquad k\geq 1\\ &=H_k^r+\frac{1}{H_1^{r-1}+\ldots+H_{k+1}^{r-1}} \end{align*}

Since $H_1^r=\frac{1}{H_1^{r-1}}$ and $H_1^{-1}=1$ according to (1), we obtain \begin{align*} H_1^r=1\qquad\forall r\geq 0 \end{align*} and we get \begin{align*} L_1=1 \end{align*}

We can now write a recurrence relation for $L_k$ \begin{align*} L_1&=1\\ L_{k+1}&=L_k+\frac{1}{L_1+\ldots+L_{k+1}} \qquad k\geq 0\tag{4} \end{align*} It's convenient to introduce \begin{align*} &S_0:= 0\\ &S_k:=L_1+\ldots+L_k\qquad k\geq 1 \end{align*}

With the help of $S_k$ we can rewrite the recurrence relation (4) as \begin{align*} L_{k+1}=L_k+\frac{1}{S_k+L_{k+1}} \end{align*}

We observe \begin{align*} L_{k+1}(S_k+L_{k+1})&=L_kS_k+L_kL_{k+1}+1\\ L_{k+1}^2+L_{k+1}(S_k-L_k)-(L_kS_k+1)&=0\\ L_{k+1}^2+L_{k+1}S_{k-1}-(L_kS_k+1)&=0\\ \end{align*} and get \begin{align*} L_{k+1}=\frac{1}{2}\left(-S_{k-1}\pm\sqrt{S_{k-1}^2+4L_kS_k+4}\right)\qquad k\geq 1 \end{align*}

Since $L_k$ is positive, we conclude with following quadratic recurrence relation: \begin{align*} L_1&=1,\quad S_0=0,\\ S_k&= L_1+\ldots+L_k \qquad &\tag{5}\\ L_k&=\frac{1}{2}\left(-S_{k-1}+\sqrt{S_{k-1}^2+4L_kS_k+4}\right)\qquad &k\geq 1 \end{align*}

With the help of (5) we successively find

\begin{align*} L_1&=1\\ L_2&=\sqrt{2}\approx 1.4142\\ L_3&=\frac{1}{2}\left(-1+\sqrt{13+4\sqrt{2}}\right)\approx1.65967\\ L_4&=\frac{1}{2}\left(-(1+\sqrt{2})+\sqrt{19+4\sqrt{2}+2\sqrt{2}\sqrt{13+4\sqrt{2}}}\right)\approx1.8291\\ &\ldots \end{align*}

We can equivalently express (5) using $S_k$ only.

We get from \begin{align*} S_{k}&=S_{k-1}+L_k\qquad &k\geq 1\\ L_{k+1}&=L_k+\frac{1}{S_k+L_{k+1}}\\ \end{align*} the equation \begin{align*} (S_{k+1}-S_k)&=(S_k-S_{k-1})+\frac{1}{S_k+(S_{k+1}-S_k)}\\ S_{k+1}-2S_k+S_{k-1}&=\frac{1}{S_{k+1}}\\ \end{align*} and finally \begin{align*} &S_0=0,\quad S_1=1,\\ &S_{k+1}(S_{k+1}-2S_k+S_{k-1})=1\\ \text{resp.}&\\ &S_{k+1}=\frac{1}{2}\left(2S_k-S_{k-1}+\sqrt{4S_k^2-4S_kS_{k-1}+S_{k-1}^2+4}\right) \end{align*}

The first values $S_k$, $1 \leq k \leq 4$ give in accordance with the numerical values from $L_k$ above \begin{align*} S_0=0,S_1=1,S_1\approx 2.41421,S_3\approx 4.07389,S_4\approx 5.90298 \end{align*}

Note: As far as I know, there is no systematic theory to solve quadratic recurrence relations similar to the theory we have for linear recurrences. I don't see a fruitful specific Ansatz to obtain a closed formula for $L_k$.

Observe that the expressions of $L_k$ grow in complexity with increasing $k$. Maybe in papers like this one you could find hints for studying some more aspects of quadratic recurrence relations. I don't adress the periodicity within this paper, but instead the fact that iterative solutions of quadratic equations are analysed.