Divisibility of Numerator of Sum of Sequence of Reciprocals/Lemma

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Lemma for Divisibility of Numerator of Sum of Sequence of Reciprocals

Let $n \in \Z_{>0}$ be a (strictly) positive integer.


Then:

$\ds \sum_{k \mathop = 1}^n \dfrac {\paren {-1}^{k - 1} } k \dbinom n k = \sum_{k \mathop = 1}^n \dfrac 1 k$

where $\dbinom n k$ denotes a binomial coefficient.


Proof

Expanding the summation:

Let:

\(\ds \map f n\) \(=\) \(\ds \sum_{k \mathop = 1}^n \dfrac {\paren {-1}^{k - 1} } k \dbinom n k\)
\(\ds \) \(=\) \(\ds n - \dfrac 1 2 \dbinom n 2 + \dfrac 1 3 \dbinom n 3 - \cdots + \dfrac {\paren {-1}^{n - 2} } {n - 1} \dbinom n {n - 1} + \dfrac {\paren {-1}^{n - 1} } n\)

Then we have:

\(\ds \map f {n - 1}\) \(=\) \(\ds \sum_{k \mathop = 1}^{n - 1} \dfrac {\paren {-1}^{k - 1} } k \dbinom {n - 1} k\)
\(\ds \) \(=\) \(\ds \paren {n - 1} - \dfrac 1 2 \dbinom {n - 1} 2 + \dfrac 1 3 \dbinom {n - 1} 3 - \cdots + \dfrac {\paren {-1}^{n - 3} } {n - 2} \dbinom {n - 1} {n - 2} + \dfrac {\paren {-1}^{n - 2} } {n - 1}\)
\(\ds \leadsto \ \ \) \(\ds \map f n - \map f {n - 1}\) \(=\) \(\ds \sum_{k \mathop = 1}^n \dfrac {\paren {-1}^{k - 1} } k \dbinom n k - \sum_{k \mathop = 1}^{n - 1} \dfrac {\paren {-1}^{k - 1} } k \dbinom {n - 1} k\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^{n - 1} \dfrac {\paren {-1}^{k - 1} } k \paren {\dbinom n k - \dbinom {n - 1} k} + \dfrac {\paren {-1}^{n - 1} } n\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^{n - 1} \dfrac {\paren {-1}^{k - 1} } k \dbinom {n - 1} {k - 1} + \dfrac {\paren {-1}^{n - 1} } n\) Pascal's Rule


This needs considerable tedious hard slog to complete it.
In particular: working on it, might be worth using induction
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The proof proceeds by induction.

For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:

$\ds \sum_{k \mathop = 1}^n \dfrac {\paren {-1}^{k - 1} } k \dbinom n k = \sum_{k \mathop = 1}^n \dfrac 1 k$


Basis for the Induction

$\map P 1$ is the case:

\(\ds \sum_{k \mathop = 1}^1 \dfrac {\paren {-1}^{k - 1} } k \dbinom 1 k\) \(=\) \(\ds \dfrac {\paren {-1}^0} 1 \dbinom 1 1\)
\(\ds \) \(=\) \(\ds 1\) Binomial Coefficient with One or Binomial Coefficient with Self
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^1 \dfrac 1 k\)

Thus $\map P 1$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.


So this is the induction hypothesis:

$\ds \sum_{k \mathop = 1}^r \dfrac {\paren {-1}^{k - 1} } k \dbinom r k = \sum_{k \mathop = 1}^r \dfrac 1 k$


from which it is to be shown that:

$\ds \sum_{k \mathop = 1}^{r + 1} \dfrac {\paren {-1}^{k - 1} } k \dbinom {r + 1} k = \sum_{k \mathop = 1}^{r + 1} \dfrac 1 k$


Induction Step

This is the induction step:

\(\ds \sum_{k \mathop = 1}^{r + 1} \dfrac {\paren {-1}^{k - 1} } k \dbinom {r + 1} k\) \(=\) \(\ds \sum_{k \mathop = 1}^r \dfrac {\paren {-1}^{k - 1} } k \dbinom {r + 1} k + \dfrac {\paren {-1}^r} {r + 1} \dbinom {r + 1} {r + 1}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^r \dfrac {\paren {-1}^{k - 1} } k \dbinom {r + 1} k + \dfrac {\paren {-1}^r} {r + 1}\) Binomial Coefficient with Self
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^r \dfrac {\paren {-1}^{k - 1} } k \paren {\dbinom r k + \dbinom r {k - 1} } + \dfrac {\paren {-1}^r} {r + 1}\) Pascal's Rule
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^r \dfrac {\paren {-1}^{k - 1} } k \dbinom r k + \sum_{k \mathop = 1}^r \dfrac {\paren {-1}^{k - 1} } k \dbinom r {k - 1} + \dfrac {\paren {-1}^r} {r + 1}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^r \dfrac 1 k + \sum_{k \mathop = 1}^r \dfrac {\paren {-1}^{k - 1} } k \dbinom r {k - 1} + \dfrac {\paren {-1}^r} {r + 1}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^r \dfrac 1 k + \sum_{k \mathop = 0}^{r - 1} \dfrac {\paren {-1}^k} {k + 1} \dbinom r k + \dfrac {\paren {-1}^r} {r + 1}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^r \dfrac 1 k + \sum_{k \mathop = 0}^r \dfrac {\paren {-1}^k} {k + 1} \dbinom r k\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^r \dfrac 1 k + \sum_{k \mathop = 0}^r \dfrac {\paren {-1}^k} {k + 1} \paren { \dfrac {k + 1} {r + 1} \dbinom {r + 1} {k + 1} }\) Factors of Binomial Coefficient
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^r \dfrac 1 k + \dfrac 1 {r + 1} \sum_{k \mathop = 0}^r \paren {-1}^k \dbinom {r + 1} {k + 1}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^r \dfrac 1 k + \dfrac 1 {r + 1} \sum_{k \mathop = 1}^{r + 1} \paren {-1}^{k - 1} \dbinom {r + 1} k\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^r \dfrac 1 k + \dfrac 1 {r + 1} \paren { - \sum_{k \mathop = 0}^{r + 1} \paren {-1}^k \dbinom {r + 1} k + \dbinom {r + 1} 0 }\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^r \dfrac 1 k + \dfrac 1 {r + 1}\) Alternating Sum and Difference of Binomial Coefficients for Given n, Binomial Coefficient with Zero
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^{r + 1} \dfrac 1 k\)


So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\ds \forall n \in \Z_{>0}: \sum_{k \mathop = 1}^n \dfrac {\paren {-1}^{k - 1} } k \dbinom n k = \sum_{k \mathop = 1}^n \dfrac 1 k$


$\blacksquare$

Sources

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