0

I tried simplifying the left-hand side(the harmonic series summation formula) and came up with the right-hand side by finding $\frac{1}{r} - \frac{1}{2r} = \frac{1}{2r}$ and then evaluating by plugging in values of r and cancelling out the terms to find the two summations on the RHS below.

$\sum_{r=1}^{n}(\frac{1}{2r}) =\sum_{r=1}^{n/2}( \frac{1}{2r-1}) - \sum_{r=\frac{n}{2}+1}^{n} (\frac{1}{2r})$

Is there a way to simplify the right-hand side any further or is this a dead end?

I just think that there should be a formula for a summation to some value of n. Also not looking to find any formula for the infinite summation of the harmonic series.

Sorry this is so muddled :s

MadCanadian
  • 1

2 Answers2

0

I don't know what you're trying to achieve. The usual formula for the partial sums of the harmonic series is

$$ \sum_{r=1}^n \frac{1}{r} = \Psi(n+1) + \gamma $$

There are many identities involving $\Psi$. Your formula seems to be related to the identity

$$ \Psi(2n+1) = \frac{\Psi(n) + \Psi(n+1)}{2} + \ln(2) + \frac{1}{2n}$$

Robert Israel
  • 448,999
0

What you have written is $\frac12H_n$ where $H_n$ denotes the $n$th harmonic number (sum up to $n$ terms of $\frac1r$). One closed form for $H_n$ is the following $$H_n=\frac{\binom{n+(n+1)!}{n}-1}{(n+1)!}-(n+1)\left\lfloor\frac{\binom{n+(n+1)!}{n}-1}{(n+1)\cdot(n+1)!}\right\rfloor$$

Peter Foreman
  • 19,947