1
$\begingroup$

I know that the harmonic number $H_a ^{(b)}$ is $$\sum_{n=1}^a \frac{1}{n^b}$$ I was wondering if, for the generalized alternating harmonic number $\bar H_a^{(b)}$, there was a closed formula. For example, when $b$ is 1, $$\sum_{n=1}^a \frac{(-1)^{n-1}}{n^b}$$ is conditionally convergent to $\ln(2)$ because of the taylor series representation of $\ln(1+x)$. I'm asking if similar techniques are possible to find the value with a variable $b$.

$\endgroup$
2
$\begingroup$

For $x > 1$, $$ \sum_{n=1}^\infty \frac{1}{n^x} = \zeta(x) $$ and
$$ \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^x} = (1 - 2^{1-x})\zeta(x) \text{,} $$ where $\zeta(x)$ is the Riemann zeta function. The fact you cite for $x = 1$ follows from the "coincidence", for $|x|<1$, $$ \ln(1+x) = \sum_{n = 1}^\infty (-1)^{n+1} \frac{x^k}{k} \text{,} $$ which we may evaluate at $x = 1$ to get the alternating harmonic series. The corresponding coincidence for $x = 2$ is the dilogarithm: $$ \mathrm{Li}_2(x) = \sum_{n=1}^\infty \frac{x^n}{n^2} \text{,} $$ which we may evaluate ate $x = -1$ to get the alternating series with $b=2$ about which you ask. More generally, the polylogarithms are the functions for the series you ask about. $$ \mathrm{Li}_b(x) = \sum_{n=1}^\infty \frac{x^n}{n^b} \text{,} $$ which we evaluate at $x = -1$ to get the alternating series.

$\endgroup$
  • $\begingroup$ Thank you. I had read about a correlation between harmonics and polygamma functions so it’s cool that there is a similar correlation for the alternating harmonic too. $\endgroup$ – Ryan Shesler Apr 24 at 11:45
1
$\begingroup$

$$\sum_{n=1}^a \frac{(-1)^{n-1}}{n^b}=2^{-b} \left(\left(2^b-2\right) \zeta (b)+(-1)^a \left(\zeta \left(b,\frac{a+2}{2}\right)-\zeta \left(b,\frac{a+1}{2}\right)\right)\right)$$ $$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^b}=2^{-b} \left(2^b-2\right) \zeta (b)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.