Summation to n of Square of kth Harmonic Number

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Theorem

$\ds \sum_{k \mathop = 1}^n {H_k}^2 = \paren {n + 1} {H_n}^2 - \paren {2 n + 1} H_n + 2 n$

where $H_k$ denotes the $k$th harmonic number.


Proof

Consider:

\(\ds \sum_{k \mathop = 1}^{n - 1} \dfrac k {k + 1}\) \(=\) \(\ds \sum_{k \mathop = 1}^{n - 1} \dfrac {\paren {k + 1} - 1} {k + 1}\) factorizing
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^{n - 1} 1 - \sum_{k \mathop = 1}^{n - 1} \dfrac 1 {k + 1}\) Sum of Summations equals Summation of Sum
\(\ds \) \(=\) \(\ds n - 1 - \sum_{k \mathop = 1}^{n - 1} \dfrac 1 {k + 1}\) Summation of Unity over Elements
\(\ds \) \(=\) \(\ds n - 1 - \paren {H_n - 1}\) Harmonic Number and compensate for $\dfrac 1 1$ term
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds n - H_n\) simplification





Using Summation by Parts:

\(\ds \sum_{k \mathop = 1}^n {H_k}^2\) \(=\) \(\ds H_n \sum_{k \mathop = 1}^n H_k - \sum_{k \mathop = 1}^{n - 1} \paren {\paren {\sum_{i \mathop = 1}^k {H_i} } \paren {H_{k + 1} - H_k} }\)
\(\ds \) \(=\) \(\ds H_n \paren {\paren {n + 1} H_n - n} - \sum_{k \mathop = 1}^{n - 1} \paren {\paren {k + 1} H_k - k} \paren {H_{k + 1} - H_k}\) Sum of Sequence of Harmonic Numbers twice
\(\ds \) \(=\) \(\ds \paren {n + 1} {H_n}^2 - n H_n - \sum_{k \mathop = 1}^{n - 1} \dfrac {\paren {k + 1} H_k - k} {k + 1}\) $H_{k + 1} - H_k = \dfrac 1 {k+1}$
\(\ds \) \(=\) \(\ds \paren {n + 1} {H_n}^2 - n H_n - \sum_{k \mathop = 1}^{n - 1} H_k + \sum_{k \mathop = 1}^{n - 1} \dfrac k {k + 1}\) Sum of Summations equals Summation of Sum
\(\ds \) \(=\) \(\ds \paren {n + 1} {H_n}^2 - n H_n - \paren { \sum_{k \mathop = 1}^n H_k - H_n} + \sum_{k \mathop = 1}^{n - 1} \dfrac k {k + 1}\) increase range of summation and compensate
\(\ds \) \(=\) \(\ds \paren {n + 1} {H_n}^2 - n H_n - \paren {\paren {n + 1} H_n - n - H_n } + \sum_{k \mathop = 1}^{n - 1} \dfrac k {k + 1}\) Sum of Sequence of Harmonic Numbers
\(\ds \) \(=\) \(\ds \paren {n + 1} {H_n}^2 - n H_n - \paren {n + 1} H_n + n + H_n + n - H_n\) $(1)$
\(\ds \) \(=\) \(\ds \paren {n + 1} {H_n}^2 - \paren {2 n + 1} H_n + 2 n\) simplification

$\blacksquare$


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