Closed form for $1+\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}$?

Question

Background

When I attempt questions in real analysis, frequently I encounter the following expression like harmonic series:

$$H_n=1+\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}$$

To my surprise when I arrive at something like this ,the solution book does not give me a closed-form of this expression.

Question

So my question is a simple one. What can you think of for this series? And besides the fact that it does not converge is it related to any mathematics that you know? And most importantly, does it have a closed-form?

Answer 1

The harmonic series diverges, which is somewhat surprising because each term tends to zero as $n\to\infty$. However, the problem is that each term does not tend to zero fast enough.

A function which generalises this series is the Riemann zeta function $$\zeta(s)=\sum_{n=1}^\infty\frac{1}{n^s}$$ which converges for $\text{Re}(s)>1$. For example,

$$\zeta(2)=\frac{\pi^2}{6}.$$

Euler gave a general formula for $\zeta(2n)$ in terms of the Bernoulli numbers, $$\zeta(2n)=\frac{(-1)^{n+1}B_{2n}(2\pi)^{2n}}{2(2n)!},$$ but no closed form is known yet for odd arguments.

It can also be analytically continued to $\mathbb{C}\setminus\{1\}$. There is a singularity (pole of order 1) at $s=1$.

Zeros of this function lie on the negative even integers (the trivial zeros), whereas the non trivial (purely complex) zeros are all known to lie in the critical strip. In fact it is hypothesised that they all lie on the critical line $s=1/2+it$, which you may know is called the Riemann Hypothesis. If true, this has big implications concerning the distribution of the primes, and many conjectured theorems in Analytic Number Theory, since $\zeta(s)$ is related to the primes by Euler's product formula, $$\zeta(s)=\prod_{p\text{ prime}}\left(1-p^{-s}\right)^{-1}.$$

Coming back to "$\zeta(1)$" for a moment, as mentioned in the other answer, it pops up in the definition of Euler's gamma constant,

$$\gamma=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k}-\log n\approx 0.57721566490...$$

It is not yet known whether $\gamma$ is irrational, unlike $\zeta(2n)$, and also $\zeta(3)$ which was proved to be irrational by Apéry.

It may also be of interest to note that $$H_n=\sum_{k=1}^n\frac{1}{k}=\Psi(n+1)+\gamma,$$ where $\Psi$ is the Digamma function.

Another remarkable "closed form" is given by

$$H_n = \frac{\binom{(n+1)!+n}{n}-1}{(n+1)!}-(n+1)\Biggl\lfloor \frac{\binom{(n+1)!+n}{n}-1}{(n+1)(n+1)!}\Biggr\rfloor,$$

as stated by @nczksv in the "duplicate" answer.

Answer 2

If I remember correctly, there is a sort of closed form:

$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{n} = \sum_{k=1}^n \frac{1}{k} = \ln(n) + \gamma + \epsilon_n$$

$\epsilon_n$ here is an error constant proportional to $1/2n$, and thus $\epsilon_n \to 0$ as $n \to \infty$.

$\gamma$ denotes the Euler-Mascheroni constant, defined by the limiting difference between the natural logarithm and the harmonic series, i.e.

$$\gamma =\lim _{n\to \infty }\left(-\ln n+\sum _{k=1}^{n}{\frac {1}{k}} \right)$$

Granted, I don't think this is really in line with what you want because ... well, it's sort of like a self-referential thing. "Oh, the harmonic series is given by a function, an error constant, and this special constant that comes from the difference from the series and that other function under certain conditions." It's mostly a personal thing so it doesn't mesh quite well with me?

Beyond that I don't really have anything to offer - specifically with relations of the series to mathematics I know - that I wouldn't just be regurgitating from Wikipedia. Weirdly it hasn't popped up much in my coursework thus far.

Answer 3

It diverges because, if you look at the partial sums, $s_{2n}-s_n=\frac1{2n}+\frac1{2n-1}+\cdots+\frac1{n+1}\ge n\cdot \frac1{2n}=\frac12\,,\forall n$. Thus it isn't Cauchy.