General term formula of series 1/1 + 1/2 + 1/3 ... +1/n

Question

any hint how to resolve

$$f(n) = \frac 11 + \frac 12 +\frac 13 + \dots + \frac 1n$$

What I'm trying to do is to find connection between

$$f(n),\,f(n+1)$$ different of

$$f(n+1) = f(n)+\frac 1{n+1}$$

So I could create system.

Answer 1

This is the harmonic series. It has no closed-form solution to the extent of my knowledge. In the limit, it is divergent (meaning that it sums to infinity). I can't think of any good way to recast the equation over a common denominator (which would be $n!$).

Answer 2

Since the comments above the solution of the problem

$f(n) = 1/1 + 1/2 +1/3 + ... + 1/n$

is using Harmonics Numbers:

$f(n)=H_n$ the $n$-th harmonic number.

Answer 3

$$\Gamma(x+1)=x\Gamma(x)$$ $$\Gamma'(x+1)=\Gamma(x)+x\Gamma'(x)$$ $$\frac{1}{x}=\frac{\Gamma'(x+1)}{x\Gamma(x)}-\frac{\Gamma'(x)}{\Gamma(x)}=\frac{\Gamma'(x+1)}{\Gamma(x+1)}-\frac{\Gamma'(x)}{\Gamma(x)}$$ $$\sum_{i=1}^n\frac{1}{i}=\frac{\Gamma'(n+1)}{n!}-\Gamma'(1)=\frac{\Gamma'(n+1)}{n!}+\gamma$$ where $\gamma$ is the Euler constant.

Answer 4

I can give you a good approximation if you would prefer:

$$\ln(n+1)\le\sum_{i=1}^n\frac1i\le\ln(n)+1$$

This is a rather tight upper limit and lower limit you can use to approximate your answer.

One could also note that

$$\sum_{i=1}^n\frac1i=\int_0^1\sum_{i=0}^{n-1}x^i\ dx=\int_0^1\frac{1-x^n}{1-x}\ dx$$

We also have the Euler-Maclaurin expansion:

$$\sum_{i=1}^n\frac1i=\ln(n)+\gamma+\frac1n-\frac1{4n^2}+\mathcal O(n^{-4})$$