Is there a general formula for harmonic number at present? New to stackexchange.
Known harmonic numbers are defined as
$$ H_n= \sum _ {i = 1}^n\frac {1} {i}$$
Is the series above similar to the general term formula of  $ \sum _ {i = 1}^n i= n (n+1)/2$?


The method is the same as for $\sum_{j \le n} j^r$ but the result isn't a polynomial, we only have an asymptotic expansion.

Let $$f(z) = \frac1z - \log(z+1)+\log(z)=\frac1z - \log(1+\frac1{z})=F(\frac1z)$$
$F(s)= s-\log(1+s)$ is analytic for $|s|<1$ and $F(0)=F'(0)=0$. Thus, by induction there are some coefficients $c_k$ such that for all $K$, the first $K+1$ derivatives of $$F_K(s) = \sum_{k=1}^K c_k s^k (1-(\frac1{s+1})^k)$$ agree with that of $F$.
Whence $$\sum_{j> n} f(j)= \sum_{j> n} (F_K(1/j)+O(j^{-K-2}))= \sum_{k=1}^K c_k \sum_{j> n}(\frac1{j^k}-\frac1{(j+1)^k})+O(\sum_{j>n} j^{-K-2})$$
$$ =  \sum_{k=1}^K \frac{c_k }{(n+1)^k} + O(n^{-K-1})$$ From which we get
$$\sum_{j \le n} \frac1j = \sum_{j\le n} (\log(j+1)-\log(j)) + \sum_j f(j) - \sum_{j> n} f(j)$$
$$ = \log(n+1)+\gamma-\sum_{k=1}^K  \frac{c_k}{(n+1)^k} + O(n^{-K-1})$$
