Convergence of $\left( \frac{1}{1} \right)^2+\left( \frac{1}{2}+\frac{1}{3} \right)^2+\cdots$

Does the following series converge? If yes, what is its value in simplest form?

$$\left( \frac{1}{1} \right)^2+\left( \frac{1}{2}+\frac{1}{3} \right)^2+\left( \frac{1}{4}+\frac{1}{5}+\frac{1}{6} \right)^2+\left( \frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10} \right)^2+\dots$$

I have no idea how to start. Any hint would be really appreciated. THANKS!

Solution 1:

Notice, that:

$$\left(\frac{1}{1}\right)^2+\left(\frac{1}{2} + \frac{1}{3}\right)^2 + \dots<\left(\frac{1}{1}\right)^2+\left(\frac{2}{2}\right)^2+\left(\frac{3}{4}\right)^2+\left(\frac{4}{7}\right)^2=2+\sum_{n=2}^{\infty}\left(\frac{n}{\frac{n(n-1)}{2}+1}\right)^2<2+\sum_{n=2}^{\infty}\left(\frac{2}{n-1}\right)^2$$

Since it's bounded by converging series, it's convergent itself.

Solution 2:

Approximation with Euler-Maclaurin Sum Formula

The Euler-Maclaurin Sum Formula gives the fairly well-known asymptotic series for the Harmonic Numbers $$ \begin{align} H_n &=\gamma+\log(n)+\frac1{2n}-\frac1{12n^2}+\frac1{120n^4}\\[3pt] &-\frac1{252n^6}+\frac1{240n^8}-\frac1{132n^{10}}+O\!\left(\frac1{n^{12}}\right)\tag1 \end{align} $$ where $\gamma$ is the Euler-Mascheroni constant. $\gamma$ doesn't actually come from the Euler-Maclaurin Sum Formula, but is defined as $\lim\limits_{n\to\infty}(H_n-\log(n))$.

Substituting into $(1)$ and expanding Taylor series gives $$ \begin{align} \left(H_{n(n+1)/2}-H_{n(n-1)/2}\right)^2 &=\frac4{n^2}-\frac{16}{3n^4}+\frac{32}{45n^6}+\frac{1424}{315n^8}\\[3pt] &+\frac{3392}{1575n^{10}}-\frac{112912}{10395n^{12}}+O\!\left(\frac1{n^{14}}\right)\tag2 \end{align} $$ Equation $(2)$ not only shows that the series converges, but applying the Euler-Maclaurin Sum Formula to $(2)$ results in $$ \begin{align} &\sum_{k=1}^n\left(H_{k(k+1)/2}-H_{k(k-1)/2}\right)^2\\ &=C-\frac4n+\frac2{n^2}+\frac{10}{9n^3}-\frac8{3n^4}+\frac{398}{225n^5}+\frac{16}{45n^6}-\frac{4378}{2205n^7}\\[3pt] &+\frac{712}{315n^8}-\frac{22718}{14175n^9}+\frac{1696}{1575n^{10}}+\frac{138}{4235n^{11}}-\frac{56456}{10395n^{12}}+O\!\left(\frac1{n^{13}}\right)\tag3 \end{align} $$ Plugging $n=200$ into $(3)$ gives $$ \bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^\infty\left(H_{k(k+1)/2}-H_{k(k-1)/2}\right)^2=3.170466061427153684757796531}\tag4 $$ As a check, $n=300$ gives the same result.

Solution 3:

Since for moderately large values of $n$ we have $$ H_n \approx \log(n)+\gamma+\frac{1}{2n}-\frac{1}{12n^2} $$ we also have $$ H_{n(n+1)/2}-H_{n(n-1)/2} \approx \frac{2}{n}-\frac{4}{3n^3}$$ and $$ \sum_{n\geq 1}\left[H_{n(n+1)/2}-H_{n(n-1)/2}\right]^2 \approx 1+\sum_{n\geq 2}\left(\frac{2}{n}-\frac{4}{3n^3}\right)^2=1+\frac{2 \left(-1890+2835 \pi ^2-252 \pi ^4+8 \pi ^6\right)}{8505} $$ is finite and approximately equal to $\color{green}{3.17}151$. I doubt there is a simple closed form.

Solution 4:

Suggestion.

This is merely a suggestion that is too large to fit in a comment. Euler's integral formula for the harmonic numbers, $H_n=\int_0^1\frac{1-x^n}{1-x}\ \mathrm{d}x$, gives us a formula for the series as $a\to1^-$:

$$S=4\sum_{n=1}^{\infty}\left[f_a(n(n+1))-f_a(n(n-1))\right]^2$$

where $f_a(x)=\int_{0}^{a}\frac{t^{x+1}}{1-t^{2}}\ \mathrm{d}t$. The function, $f_a$, can also be given in terms of the hypergeometric function, as $\frac{a^{x+2}}{x+2}\cdot{_2F}_1\left(1,\frac{x}2+1;\frac{x}2+2;a^{2}\right)$.

Thus, it could potentially aid in finding an analytic solution to $S$ (or showing whether that would be impossible) by finding a closed form for the square of the difference of hypergeometric functions, $\left[f_a(n(n+1))-f_a(n(n-1))\right]^2$. However, no such solution has leapt out to me so far.