# Limit of $a_n = \frac{1}{1^3\cdot 1}+\frac{1}{1^3\cdot 2+2^3\cdot 1}+\cdots+\frac{1}{1^3\cdot n+2^3\cdot (n-1)+\cdots+n^3\cdot 1}$

Let

$$a_n = \frac{1}{1^3\cdot 1}+\frac{1}{1^3\cdot 2+2^3\cdot 1} + \cdots +\frac{1}{1^3\cdot n+2^3\cdot (n-1)+\cdots+n^3\cdot 1}$$

Does this sequence converge to a simple number?

My thought was to compute each denominator:

$$1^3\cdot n+2^3\cdot (n-1)+\cdots+n^3\cdot 1=(n+1)\sum_{k=1}^n k^3-\sum_{k=1}^nk^4$$

and this are known, I find $$\frac{1}{60}n(n+1)(2n+1)(3n^2+6n+1)$$. But can we find maybe a closed formula for $$a_n$$ after this?

• it's certainly clear it converges, perhaps using partial fractions is the way to many telescoping series – gt6989b Feb 10 at 20:31
• You have the explicit form for the general term of the series. So, try partial fraction expansion next and see if you can find a closed form. – Mark Viola Feb 10 at 20:37
• @MarkViola, I find $30(\frac{1}{k}+\frac{1}{k+1}+\frac{1}{k+2}-\frac{9(k+1)}{3k^2+6k+1})$, but I cant telescope this. – user748957 Feb 10 at 20:41
• Well telescoping series are not the only ones for which close forms can be found. – Mark Viola Feb 10 at 20:44
• The sum is 1.134103506, obviously it should be less than $\zeta(3)$ – Oldboy Feb 11 at 7:52

Your sum is: $$\sum_{k=1}^{n}\frac{1}{\left(1^{3}\cdot k\right)+\left(2^{3}\cdot\left(k-1\right)\right)+...+\left(k^{3}\cdot\left(1\right)\right)}=\sum_{k=1}^{n}\frac{1}{\color{red}{\sum_{m=1}^{k}m^{3}\cdot\left(k+1-m\right)}}$$ For the red part we have: $$\sum_{m=1}^{k}m^{3}\cdot\left(k+1-m\right)=\color{blue}{\left(k+1\right)\sum_{m=1}^{k}m^{3}}-\color{green}{\sum_{m=1}^{k}m^{4}}$$ Using Faulhaber's formula follows:

$$\color{blue}{\left(k+1\right)\cdot\frac{k^{4}+2k^{3}+k^{2}}{4}}-\color{green}{\frac{k\left(k+1\right)\left(2k+1\right)\left(3k^{2}+3k-1\right)}{30}}$$ Replace this relation in the main sum:

$$\sum_{k=1}^{n}\frac{1}{\left(k+1\right)\cdot\frac{k^{4}+2k^{3}+k^{2}}{4}-\frac{k\left(k+1\right)\left(2k+1\right)\left(3k^{2}+3k-1\right)}{30}}$$$$=60\sum_{k=1}^{n}\frac{1}{15\left(k+1\right)\cdot\left(k^{4}+2k^{3}+k^{2}\right)-2k\left(k+1\right)\left(2k+1\right)\left(3k^{2}+3k-1\right)}$$$$=60\sum_{k=1}^{n}\frac{1}{3k^{5}+15k^{4}+25k^{3}+15k^{2}+2k}$$

Notice that : $$3k^{5}+15k^{4}+25k^{3}+15k^{2}+2k$$$$=k\left(3k^{4}+15k^{3}+25k^{2}+15k+2\right)$$

Clearly one of the roots is $$k=0$$.

Assume rational roots of the other part are in the form $$\frac{p}{q}$$ where $$p,q∈ℤ$$ and $$q≠0$$, also assume this fraction is in the simplest form ,using rational root theorem implies $$p$$ must divide $$2$$ and $$q$$ must divide $$3$$, so the whole fractions with these assumptions are:

$$\pm1 , \pm2 ,\pm\frac{1}{3} , \pm\frac{2}{3}$$

Checking them implies $$-1,-2$$ are two integer roots of the equation.

So we apply what we derived: $$=60\sum_{k=1}^{n}\frac{1} {k\left(k+1\right)\left(k+2\right)\left(3k^{2}+6k+1\right)}$$ Using partial fraction decomposition we have:

$$=60\left[\sum_{k=1}^{n}\frac{1}{2k}+\sum_{k=1}^{n}\frac{1}{2\left(k+2\right)}+\sum_{k=1}^{n}\frac{1}{2\left(k+1\right)}+\sum_{k=1}^{n}-\frac{9}{2}\cdot\frac{k+1}{3k^{2}+6k+1}\right]$$

$$=30\left[\color{blue}{\sum_{k=1}^{n}\frac{1}{k}}+\color{red}{\sum_{k=1}^{n}\frac{1}{k+2}}+\color{green}{\sum_{k=1}^{n}\frac{1}{k+1}}-\color{orange}{9\sum_{k=1}^{n}\frac{k+1}{3k^{2}+6k+1}}\right]$$

For calculating the orange part we have:

$$9\sum_{k=1}^{n}\frac{k+1}{3k^{2}+6k+1}$$$$=\frac{9}{2}\left[\sum_{k=1}^{n}\frac{1}{3k+3+\sqrt{6}}+\sum_{k=1}^{n}\frac{1}{3k+3-\sqrt{6}}\right]$$

$$=\frac{3}{2}\left[\sum_{k=1}^{n}\frac{1}{k+1+\frac{\sqrt{6}}{3}}+\sum_{k=1}^{n}\frac{1}{k+1-\frac{\sqrt{6}}{3}}\right]$$

Setting $$k+1+\frac{\sqrt{6}}{3} \mapsto k$$ and $$k+1-\frac{\sqrt{6}}{3} \mapsto k'$$ yields:

$$=\frac{3}{2}\left[\sum_{k=2+\sqrt{\frac{2}{3}}}^{n+1+\sqrt{\frac{2}{3}}}\frac{1}{k}+\sum_{k'=2-\sqrt{\frac{2}{3}}}^{n+1-\sqrt{\frac{2}{3}}}\frac{1}{k'}\right]$$$$=\frac{3}{2}\left[\sum_{k=1}^{n+1+\sqrt{\frac{2}{3}}}\frac{1}{k}-\sum_{k=1}^{1+\sqrt{\frac{2}{3}}}\frac{1}{k}+\sum_{k'=1}^{n+1-\sqrt{\frac{2}{3}}}\frac{1}{k'}-\sum_{k'=1}^{1-\sqrt{\frac{2}{3}}}\frac{1}{k'}\right]=\frac{3}{2}\left[H_{n+1+\sqrt{\frac{2}{3}}}-H_{1+\sqrt{\frac{2}{3}}}+H_{n+1-\sqrt{\frac{2}{3}}}-H_{1-\sqrt{\frac{2}{3}}}\right]$$ $$=30\left(\color{blue}{H_{n}}+\color{red}{H_{n+2}-\frac{3}{2}}+\color{green}{H_{n+1}-1}-\color{orange}{\frac{3}{2}\left[H_{n-\sqrt{\frac{2}{3}}+1}+H_{n+\sqrt{\frac{2}{3}}+1}-H_{1+\sqrt{\frac{2}{3}}}-H_{1-\sqrt{\frac{2}{3}}}\right]}\right)$$

Where $$H_n$$ is the n-th harmonic number.

And that is the closed form you where looking for.

Now you need some simple addition subtraction tricks and using the following fact: $$\lim_{n\to\infty}\left(H_{n}-\ln\left(n\right)\right)$$$$=\lim_{n\to\infty}\left(H_{n+1}-\ln\left(n+1\right)\right)$$$$=\lim_{n\to\infty}\left(H_{n+2}-\ln\left(n+2\right)\right)$$$$=\lim_{n\to\infty}\left(H_{n-\sqrt{\frac{2}{3}}+1}-\ln\left(n-\sqrt{\frac{2}{3}}+1\right)\right)$$$$=\lim_{n\to\infty}\left(H_{n+\sqrt{\frac{2}{3}}+1}-\ln\left({n+\sqrt{\frac{2}{3}}+1}\right)\right)$$$$=\gamma$$

$$\gamma$$ is Euler–Mascheroni constant.

Finally taking the limit of the relation we get:

$$=30\lim_{n\to\infty}\left(3\gamma-\frac{3}{2}(2\gamma)-\frac{5}{2}\right)$$ $$+30\lim_{n\to\infty}\ln\left(\frac{n\left(n+2\right)\left(n+1\right)}{\sqrt{\left(\left(n-\sqrt{\frac{2}{3}}+1\right)\left(n+\sqrt{\frac{2}{3}}+1\right)\right)^{3}}}\right)$$

$$-45\lim_{n\to\infty}\left(-H_{1+\sqrt{\frac{2}{3}}}-H_{1-\sqrt{\frac{2}{3}}}\right)$$

$$\simeq\bbox[5px,border:2px solid #C0A000]{1.134103506}$$

Continuing in a different way, you can express the limit in terms of the digamma function as follows: $$\mathop {\lim }\limits_{n \to + \infty } 30\left[ {\sum\limits_{k = 1}^n {\frac{1}{k}} + \sum\limits_{k = 1}^n {\frac{1}{{k + 1}}} + \sum\limits_{k = 1}^n {\frac{1}{{k + 2}}} - \frac{3}{2}\sum\limits_{k = 1}^n {\frac{1}{{k + 1 + \sqrt {2/3} }}} - \frac{3}{2}\sum\limits_{k = 1}^n {\frac{1}{{k + 1 - \sqrt {2/3} }}} } \right] \\ = \mathop {\lim }\limits_{n \to + \infty } 30\left[ {\frac{1}{2} - \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + \frac{3}{2}\sum\limits_{k = 1}^n {\left[ {\frac{1}{{k + 1}} - \frac{1}{{k + 1 + \sqrt {2/3} }}} \right]} + \frac{3}{2}\sum\limits_{k = 1}^n {\left[ {\frac{1}{{k + 1}} - \frac{1}{{k + 1 - \sqrt {2/3} }}} \right]} } \right] \\ = \mathop {\lim }\limits_{n \to + \infty } 30\left[ {6 + \frac{1}{2} - \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + \frac{3}{2}\sum\limits_{k = 0}^n {\left[ {\frac{1}{{k + 1}} - \frac{1}{{k + 1 + \sqrt {2/3} }}} \right]} + \frac{3}{2}\sum\limits_{k = 0}^n {\left[ {\frac{1}{{k + 1}} - \frac{1}{{k + 1 - \sqrt {2/3} }}} \right]} } \right] \\ = 30\left[ {6 + \frac{1}{2} + \frac{3}{2}\left( {\psi \left( {\sqrt {2/3} } \right) + \gamma } \right) + \frac{3}{2}\left( {\psi \left( { - \sqrt {2/3} } \right) + \gamma } \right)} \right] \\ = 195 + 90\gamma + 45\left( {\psi \left( {\sqrt {2/3} } \right) + \psi \left( { - \sqrt {2/3} } \right)} \right) \\ = 195 + 90\gamma + 45\pi \cot (\pi \sqrt {2/3} ) + 45\sqrt {\frac{3}{2}} + 90\psi (\sqrt {2/3} ).$$