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In the description of this question, it was established that \begin{align} \sum_{n=2}^{\infty} (\zeta(n)^{2}-1) &= \require{cancel} \cancel{ \frac{7}{4} - \zeta(2) + 2\sum_{m=2}^{\infty} \frac{H_{m-\frac{1}{m}} - H_{-\frac{1}{m}} - H_{m} }{m} } \\ &= \frac{7}{4} - \zeta(2) + 2 \sum_{m=1}^{\infty} \frac{H_{m-1- \frac{1}{m}} - H_{- \frac{1}{m}} - H_{m-1} }{m} \qquad(1) \end{align}

Edit: the original expresion was incorrect, it has now been corrected, as per Carlo Beenakker's answer to the third question. The first two questions remain as of yet unanswered.

In the answer to that very same question, Fedor Petrov showed that an integral representation can be found for this sum. If we proceed from his derivation and use this calculation, we obtain \begin{align} \sum_{n=2}^{\infty}(\zeta(n)^{2}-1) &= 1 + \int_0^1 \sum_{k=0}^\infty\frac{x^{k}}{1+x+\ldots+x^{k+1}}dx \\ &= 1 + \sum_{k=0}^{\infty} \int_{0}^{1}\frac{x^{k}}{1+x+\ldots+x^{k+1}}dx \\ &= 1 - \sum_{k=0}^{\infty} \frac{H_{- \frac{1}{k+2}}}{k+2} \\ &= 1- \sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m}.\qquad (2) \end{align}

Now, we can equate the two expressions. Note that not all terms in the first sum can be separated, as only $\sum \frac{H_{-\frac{1}{m}}}{m}$ converges as a standalone series. Grouping like terms together, we find $$\sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m} = \frac{3}{4} - \zeta(2) + 2 \sum_{m=2}^{\infty} \frac{H_{m - 1- \frac{1}{m}} - H_{m-1} }{m}. \qquad \qquad (*)$$

What I find interesting here, is that (series involving) Harmonic numbers with both negative and positive fractional arguments can be related to one another.

Questions:

1. Do identities like the $(*)$-marked equation appear in the literature?
2. Can it be shown that one side of the equation amounts to the other side, only by means of algebraic manipulations and without invoking the aforementioned integral representation?
3. I seem to have made some calculation error, because the sums don't appear to add up to the same number. Can this error be identified? Answer: this has been answered by Carlo Beenakker. The correct identity has now been established. The first two questions remain open.

In the description of this question, it was established that \begin{align} \sum_{n=2}^{\infty} (\zeta(n)^{2}-1) &= \frac{7}{4} - \zeta(2) + 2 \sum_{m=1}^{\infty} \frac{H_{m-1- \frac{1}{m}} - H_{- \frac{1}{m}} - H_{m-1} }{m} \qquad(1) \end{align}

In the answer to that very same question, Fedor Petrov showed that an integral representation can be found for this sum. If we proceed from his derivation and use this calculation, we obtain \begin{align} \sum_{n=2}^{\infty}(\zeta(n)^{2}-1) &= 1 + \int_0^1 \sum_{k=0}^\infty\frac{x^{k}}{1+x+\ldots+x^{k+1}}dx \\ &= 1 + \sum_{k=0}^{\infty} \int_{0}^{1}\frac{x^{k}}{1+x+\ldots+x^{k+1}}dx \\ &= 1 - \sum_{k=0}^{\infty} \frac{H_{- \frac{1}{k+2}}}{k+2} \\ &= 1- \sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m}.\qquad (2) \end{align}

Now, we can equate the two expressions. Note that not all terms in the first sum can be separated, as only $\sum \frac{H_{-\frac{1}{m}}}{m}$ converges as a standalone series. Grouping like terms together, we find $$\sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m} = \frac{3}{4} - \zeta(2) + 2 \sum_{m=2}^{\infty} \frac{H_{m - 1- \frac{1}{m}} - H_{m-1} }{m}. \qquad \qquad (*)$$

What I find interesting here, is that (series involving) Harmonic numbers with both negative and positive fractional arguments can be related to one another.

Questions:

1. Do identities like the $(*)$-marked equation appear in the literature?
2. Can it be shown that one side of the equation amounts to the other side, only by means of algebraic manipulations and without invoking the aforementioned integral representation?
3. I seem to have made some calculation error, because the sums don't appear to add up to the same number. Can this error be identified? Answer: this has been answered by Carlo Beenakker. The correct identity has now been established. The first two questions remain open.

In the description of this question, it was established that \begin{align} \sum_{n=2}^{\infty} (\zeta(n)^{2}-1) &= \require{cancel} \cancel{ \frac{7}{4} - \zeta(2) + 2\sum_{m=2}^{\infty} \frac{H_{m-\frac{1}{m}} - H_{-\frac{1}{m}} - H_{m} }{m} } \\ &= \frac{7}{4} - \zeta(2) + 2 \sum_{m=1}^{\infty} \frac{H_{m-1- \frac{1}{m}} - H_{- \frac{1}{m}} - H_{m-1} }{m} \qquad(1) \end{align}

Edit: the original expresion was incorrect, it has now been corrected, as per Carlo Beenakker's answer.

In the answer to that very same question, Fedor Petrov showed that an integral representation can be found for this sum. If we proceed from his derivation and use this calculation, we obtain \begin{align} \sum_{n=2}^{\infty}(\zeta(n)^{2}-1) &= 1 + \int_0^1 \sum_{k=0}^\infty\frac{x^{k}}{1+x+\ldots+x^{k+1}}dx \\ &= 1 + \sum_{k=0}^{\infty} \int_{0}^{1}\frac{x^{k}}{1+x+\ldots+x^{k+1}}dx \\ &= 1 - \sum_{k=0}^{\infty} \frac{H_{- \frac{1}{k+2}}}{k+2} \\ &= 1- \sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m}.\qquad (2) \end{align}

Now, we can equate the two expressions. Note that not all terms in the first sum can be separated, as only $\sum \frac{H_{-\frac{1}{m}}}{m}$ converges as a standalone series. Grouping like terms together, we find $$\sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m} = \frac{3}{4} - \zeta(2) + 2 \sum_{m=2}^{\infty} \frac{H_{m - 1- \frac{1}{m}} - H_{m-1} }{m}. \qquad \qquad (*)$$

What I find interesting here, is that (series involving) Harmonic numbers with both negative and positive fractional arguments can be related to one another.

Questions:

1. Do identities like the $(*)$-marked equation appear in the literature?
2. Can it be shown that one side of the equation amounts to the other side, only by means of algebraic manipulations and without invoking the aforementioned integral representation?
3. I seem to have made some calculation error, because the sums don't appear to add up to the same number. Can this error be identified? Answer: this has been answered by Carlo Beenakker. The correct identity has now been established. The first two questions remain open.

In the description of this question, it was established that \begin{align} \sum_{n=2}^{\infty} (\zeta(n)^{2}-1) &= \require{cancel} \cancel{ \frac{7}{4} - \zeta(2) + 2\sum_{m=2}^{\infty} \frac{H_{m-\frac{1}{m}} - H_{-\frac{1}{m}} - H_{m} }{m} } \\ &= \frac{7}{4} - \zeta(2) + 2 \sum_{m=1}^{\infty} \frac{H_{m-1- \frac{1}{m}} - H_{- \frac{1}{m}} - H_{m-1} }{m} \qquad(1) \end{align}

Edit: the original expresion was incorrect, it has now been corrected, as per Carlo Beenakker's answer to the third question. The first two questions remain as of yet unanswered.

In the answer to that very same question, Fedor Petrov showed that an integral representation can be found for this sum. If we proceed from his derivation and use this calculation, we obtain \begin{align} \sum_{n=2}^{\infty}(\zeta(n)^{2}-1) &= 1 + \int_0^1 \sum_{k=0}^\infty\frac{x^{k}}{1+x+\ldots+x^{k+1}}dx \\ &= 1 + \sum_{k=0}^{\infty} \int_{0}^{1}\frac{x^{k}}{1+x+\ldots+x^{k+1}}dx \\ &= 1 - \sum_{k=0}^{\infty} \frac{H_{- \frac{1}{k+2}}}{k+2} \\ &= 1- \sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m}.\qquad (2) \end{align}

Now, we can equate the two expressions. Note that not all terms in the first sum can be separated, as only $\sum \frac{H_{-\frac{1}{m}}}{m}$ converges as a standalone series. Grouping like terms together, we find $$\sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m} = \frac{3}{4} - \zeta(2) + 2 \sum_{m=2}^{\infty} \frac{H_{m - 1- \frac{1}{m}} - H_{m-1} }{m}. \qquad \qquad (*)$$

What I find interesting here, is that (series involving) Harmonic numbers with both negative and positive fractional arguments can be related to one another.

Questions:

1. Do identities like the $(*)$-marked equation appear in the literature?
2. Can it be shown that one side of the equation amounts to the other side, only by means of algebraic manipulations and without invoking the aforementioned integral representation?
3. I seem to have made some calculation error, because the sums don't appear to add up to the same number. Can this error be identified? Answer: this has been answered by Carlo Beenakker. The correct identity has now been established. The first two questions remain open.

In the description of this question, it was established that $$\sum_{n=2}^{\infty} (\zeta(n)^{2}-1) = \frac{7}{4} - \zeta(2) + 2\sum_{m=2}^{\infty} \frac{H_{m-\frac{1}{m}} - H_{-\frac{1}{m}} - H_{m} }{m} . \qquad(1)$$ In the answer to that very same question, Fedor Petrov showed that an integral representation can be found for this sum. If we proceed from his derivation and use this calculation, we obtain \begin{align} \sum_{n=2}^{\infty}(\zeta(n)^{2}-1) &= 1 + \int_0^1 \sum_{k=0}^\infty\frac{x^{k}}{1+x+\ldots+x^{k+1}}dx \\ &= 1 + \sum_{k=0}^{\infty} \int_{0}^{1}\frac{x^{k}}{1+x+\ldots+x^{k+1}}dx \\ &= 1 - \sum_{k=0}^{\infty} \frac{H_{- \frac{1}{k+2}}}{k+2} \\ &= 1- \sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m}.\qquad (2) \end{align}

Now, we can equate the two expressions. Note that not all terms in the first sum can be separated, as only $\sum \frac{H_{-\frac{1}{m}}}{m}$ converges as a standalone series. Grouping like terms together, we find $$\sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m} = \frac{3}{4} - \zeta(2) + 2 \sum_{m=2}^{\infty} \frac{H_{m - \frac{1}{m}} - H_{m} }{m}. \qquad \qquad (*)$$

What I find interesting here, is that (series involving) Harmonic numbers with both negative and positive fractional arguments can be related to one another.

Questions:

1. Do identities like the $(*)$-marked equation appear in the literature?
2. Can it be shown that one side of the equation amounts to the other side, only by means of algebraic manipulations and without invoking the aforementioned integral representation?
3. I seem to have made some calculation error, because the sums don't appear to add up to the same number. Can this error be identified?

In the description of this question, it was established that \begin{align} \sum_{n=2}^{\infty} (\zeta(n)^{2}-1) &= \require{cancel} \cancel{ \frac{7}{4} - \zeta(2) + 2\sum_{m=2}^{\infty} \frac{H_{m-\frac{1}{m}} - H_{-\frac{1}{m}} - H_{m} }{m} } \\ &= \frac{7}{4} - \zeta(2) + 2 \sum_{m=1}^{\infty} \frac{H_{m-1- \frac{1}{m}} - H_{- \frac{1}{m}} - H_{m-1} }{m} \qquad(1) \end{align}

Edit: the original expresion was incorrect, it has now been corrected, as per Carlo Beenakker's answer.

In the answer to that very same question, Fedor Petrov showed that an integral representation can be found for this sum. If we proceed from his derivation and use this calculation, we obtain \begin{align} \sum_{n=2}^{\infty}(\zeta(n)^{2}-1) &= 1 + \int_0^1 \sum_{k=0}^\infty\frac{x^{k}}{1+x+\ldots+x^{k+1}}dx \\ &= 1 + \sum_{k=0}^{\infty} \int_{0}^{1}\frac{x^{k}}{1+x+\ldots+x^{k+1}}dx \\ &= 1 - \sum_{k=0}^{\infty} \frac{H_{- \frac{1}{k+2}}}{k+2} \\ &= 1- \sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m}.\qquad (2) \end{align}

Now, we can equate the two expressions. Note that not all terms in the first sum can be separated, as only $\sum \frac{H_{-\frac{1}{m}}}{m}$ converges as a standalone series. Grouping like terms together, we find $$\sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m} = \frac{3}{4} - \zeta(2) + 2 \sum_{m=2}^{\infty} \frac{H_{m - 1- \frac{1}{m}} - H_{m-1} }{m}. \qquad \qquad (*)$$

What I find interesting here, is that (series involving) Harmonic numbers with both negative and positive fractional arguments can be related to one another.

Questions:

1. Do identities like the $(*)$-marked equation appear in the literature?
2. Can it be shown that one side of the equation amounts to the other side, only by means of algebraic manipulations and without invoking the aforementioned integral representation?
3. I seem to have made some calculation error, because the sums don't appear to add up to the same number. Can this error be identified? Answer: this has been answered by Carlo Beenakker. The correct identity has now been established. The first two questions remain open.