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
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
- 1862: Joseph Wolstenholme: On certain properties of prime numbers (Quart. J. Pure Appl. Math. Vol. 5: pp. 35 – 39)