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Question:

This follow series have simple closed form? $$\sum_{n=1}^{\infty}\dfrac{1}{n^2H_{n}}$$ where $$H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$$

I suddenly thought of this series, because I know this series $$\dfrac{1}{n^2H_{n}}\approx\dfrac{1}{n^2\ln{n}}$$ since $$\lim_{n\to\infty}\dfrac{\dfrac{1}{n^2\ln{n}}}{\dfrac{1}{n^2}}=0$$ so $$\sum_{n=2}^{\infty}\dfrac{1}{n^2\ln{n}}$$ converges. So $$\sum_{n=1}^{\infty}\dfrac{1}{n^2H_{n}}$$ converges. But this sum have closed form? such as link sum

and Infinite Series $\sum\limits_{n=1}^\infty\frac{H_{2n+1}}{n^2}$ Thank you

Simply Beautiful Art
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math110
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2 Answers2

2

This approach does not lead to closed form, but allows to obtain the required sum with the arbitrary accuracy.

$\color{brown}{\textbf{Preliminary notes}}$

Firstly, the harmonic sum can be presented via digamma function, $$H_n = \gamma+\psi_0(n+1),\tag1$$ then $$S=\lim\limits_{N\to\infty} S(N),\tag2$$ where $$S(N) = \sum_{n=1}^N\frac1{n^2(\gamma+\psi_0(n+1))}.\tag3$$

Secondly, the sum $S(N)$ reaches its limit slowly.
For example, $S(100000)\approx1.33275'48120$ and $S(2000000)\approx1.33275'55487.$

Thirdly, are known the asymptotic series $$\psi_0(1+z) = \log z - \frac1{2z} - \sum_{m=1}^\infty\frac{B_{2m}}{2mz^{2m}}.\tag4$$

This gives the motivation to estimate $S-S(N).$

$\color{brown}{\textbf{Estimaton}}$

Using two first terms of the sum in $(4),$ one can get $$n^2H_n = n^2(\log n+ \gamma) - \frac{60n^3+10n^2-1}{120n^4}+\dots = n^2(\log n+ \gamma)\left(1 - \frac{60n^3+10n^2-1}{120n^6(\log n+ \gamma)}+\dots\right),$$ $$S-S(N)=\sum_{n=N+1}^\infty\frac1{n^2(\log n + \gamma)}\sum_{j=0}^\infty\left( \frac{60n^3+10n^2-1}{120n^6(\log n + \gamma)}+\dots\right)^j,$$ $$S-S(N)\approx\sum_{n=N+1}^\infty\sum_{j=1}^\infty \frac{(60n^3+10n^2-1)^{j-1}}{120^{j-1}n^{6j-4}(\log n + \gamma)^j},\tag5$$ $$S\approx \sum_{n=1}^N\frac1{n^2(\gamma+\psi_0(n+1))} + \sum_{n=N+1}^\infty\sum_{j=1}^\infty \frac{(60n^3+10n^2-1)^{j-1}}{120^{j-1}n^{6j-4}(\log n + \gamma)^j}.\tag6$$

Calculations of formulas $(3),(5),(6)$ give: \begin{pmatrix} N & S(N) & S-S(N) & S\\ 10 & 1.30752'69686 & 0.02542'72 & 1.33295'42\\ 20 & 1.32179'08559 & 0.01100'11 & 1.33279'20\\ 50 & 1.32906'32996 & 0.00369'629 & 1.33275'959\\ 100 & 1.33111'40166 & 0.00164'234 & 1.33275'546\\ 200 & 1.33201'79513 & 0.00073'7784 & 1.33275'5735\\ 500 & 1.33249'56726 & 0.00025'9926 & 1.33275'5599\\ 1000 & 1.33263'64101 & 0.00011'9174 & 1.33275'5584\\ 2000 & 1.33270'05664'81413 & 0.00005'50143 & 1.33275'55807\\ 5000 & 1.33273'56014'83833 & 0.00001'99785 & 1.33275'55800\\ 10000 & 1.33274'62412'69296 & 0.00000'93386'5 & 1.33275'55799'2\\ 20000 & 1.33275'11959'85027 & 0.00000'43839'2 & 1.33275'55799'1\\ 50000 & 1.33275'39573'66156 & 0.00001'16225'3 & 1.33275'55799'0\\ 100000 & 1.33275'48120'15120 & 0.00000'07678'3 & 1.33275'55799'0\\ 200000 & 1.33275'52154'37624 & 0.00000'03644'57 & 1.33275'55798'95\\ 500000 & 1.33275'54432'72882 & 0.00000'01866'22 & 1.33275'55798'95\\ 1000000 & 1.33275'55146'82723 & 0.00000'00692'173 & 1.33275'55798'950\\ \end{pmatrix}

$\color{nrown}{\mathbf{Conclusions}}$

  1. $S\approx1.33275557895.$.
  2. Formula $(6)$ has increased accuracy compared to direct calculations.
  3. The accuracy of formula $(6)$ can be increased to the required by using additional summands in the sum of $(4)$.
  4. To reduce the formula $(6)$ to a closed form, we need at least a closed form for sums of the form $$\sum_{n=N+1}^\infty\frac1{n^p(\gamma+\log n)^q}.$$
Yuri Negometyanov
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1

While finding closed form is not likely, there's a way to get a good numerical value for this slowly converging series.

Turns out, there exist very sharp inequalities for harmonic numbers, found in this paper: https://www.sciencedirect.com/science/article/pii/S0723086906000168.

$$a-\ln \left(e^{1/(n+1)}-1 \right) \leq H_n < b-\ln \left(e^{1/(n+1)}-1 \right)$$

Where:

$$a=1+\ln \left(\sqrt{e}-1 \right)$$

$$b=\gamma$$

The equality case on the left is only for $n=1$.

The inequalitites can be used to estimate the tail of the series after computing several terms. Let's denote:

$$S_N=\sum_{n=1}^N \frac{1}{n^2 H_n}$$

$$T_N=\sum_{n=N+1}^\infty \frac{1}{n^2 H_n}$$

$$f(n)=\ln \left(e^{1/(n+1)}-1 \right)$$

We can write:

$$T_N<T_{aN}=\sum_{n=N+1}^\infty \frac{1}{n^2 \left(a-f(n)\right)}$$

$$T_N>T_{bN}=\sum_{n=N+1}^\infty \frac{1}{n^2 \left(b-f(n)\right)}$$

The sums on the right can be again estimated using integrals:

$$\int_N^\infty \frac{dx}{x^2~\left(a-f(x)\right)}<T_{aN}<\int_{N+1}^\infty \frac{dx}{x^2~\left(a-f(x)\right)}$$

The integrals likely don't have a closed form, however they can be evaluated numerically with good precision (after a suitable change of variable):

$$\int_N^\infty \frac{dx}{x^2~\left(a-f(x)\right)}=\int_0^{1/(N+1)} \frac{dy}{(1-y)^2 \left(a-\ln (e^y-1)\right)}$$

For example, we can take $N=100$, then we have only three correct digits by truncating the series:

$$S_{100}=\color{blue}{1.33}1114\dots$$

On the other hand, using the method above we can estimate the tail:

$$T_{a100}=0.00165\dots$$

$$T_{b100}=0.00163\dots$$

$$0.00163\dots< T_{100} < 0.00165\dots$$

So we obtain:

$$\color{blue}{1.3327}4\dots<S_{\infty}<\color{blue}{1.3327}6\dots$$

We have obtained $5$ correct digits using just $100$ terms. Taking the arithmetic mean of the two bounds, we get $6$ correct digits.

I might improve the numerical results later with Mathematica (for this answer I only used Wolfram Alpha).

Yuriy S
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