Is there a formulaic way to go from $\sum_{k=1}^{n} \frac{1}{k}$ back to $n$?

Question

Say you want to sum $g(n) = \sum_{k=1}^{n} \frac{1}{k} = L$. Is there a simple formula to go from $L$ and deduce $n$?

My attempt:

For $n = 1$, the formula is $L$.

Assume there is a formula for all $n=1..N$ say $f(L)$. Then we have $g(N+1) = g(N) + \frac{1}{N+1}$, $\frac{g(N)(N+1) + 1}{N+1} = L$. $(N+1)(L - g(N)) = 1$. No hope.

Answer 1

You can approximate $n$ from $L$. An overestimate is $h_n \approx \ln(n)+c+\frac{1}{n}$. where $c \approx 0.577$.

From here $$h_n-c \approx \ln(n)+1/n$$

But $$\frac{\ln(n+h)-\ln(n)}{h} \approx 1/n$$ for $h$ close to zero. So $$\ln(n+1)-\ln(n) \approx 1/n$$.

Thus,

$$h_n-c \approx \ln(n)+\ln(n+1)-\ln(n)$$ This is still valid as an overestimate for $n>e^{1-c}-1$ the intersection of noncontinuous $h_n$ (when it's graphed using steps) and $\ln(n+1)+c$. But we only care for $n \geq 1$ so we should be good.

Let $h_n=L$

$$e^{L-c} \approx n+1$$

$$e^{L-c}-1 \approx n$$

An underestimate to the harmonic series is:

$$L=h_n \approx \ln(n)+c$$

$$e^{L-c} \approx n$$

Now

$$e^{L-c}-1 < n < e^{L-c}$$

If you want to see how these approximation formulas are derived click on my profile, it should be one of my top questions titled "an approach to approximating the harmonic series".

Answer 2

Given that $n$ is an integer, you know that $\sum_{k=1}^n \frac{1}{k} \approx \ln n+\gamma+\frac1{2n}-O(n^{-2})$ ($\gamma$ is the Euler-Mascheroni constant) so you could probably use that.

Answer 3

As already said in answers and comments, for large values of $n$ $$ \sum_{k=1}^{n} \frac{1}{k}=H_n\approx \frac{1}{2 n}+\log(n)+\gamma$$ Now, the solution of $$\frac{1}{2 n}+\log(n)+\gamma=L$$ can be expressed using Lambert function $$n=-\frac{1}{2 W\left(-\frac{e^{\gamma -L}}{2}\right)}$$ Since we consider large values of $n$, the argument of Lambert function is small and we can use the approximation $$W(x)=\sum_{n=1}^\infty \frac{(-n)^{n-1}}{n!}x^n=x-x^2+\frac 32 x^3+\cdots$$ and then, $$n\approx e^{L-\gamma}-\frac 12 -\frac 1{8 e^{L-\gamma}}$$

For illustration purposes, let us consider $n=10$ which gives $L=\frac{7381}{2520}$; the approximation formula will give $n=9.99206$ which is not too bad.

Consider $L=10$; the above approximation would give $n\approx 12366.5$ while $H_{12366}\approx 9.9999621$ and $H_{12367}\approx 10.000043$.

Answer 4

$$\sum_{k=1}^{n} \frac{1}{k}=\psi(n+1)+\gamma$$

Where $\psi(x)$ is digamma function.

So to find $n$ you have to employ inverse digamma function.

$$n=\psi^{inverse}(L-\gamma)-1$$

Here is a way to define it in Mathematica: https://mathematica.stackexchange.com/questions/89107/inverse-of-a-digamma-polygamma-function