Is there a series approximation in terms of $n$ for the sum of the harmonic progression : $\sum_{k=0}^{n}\frac{1}{1+ak}$? When $a=1$ the sum is given by $ H_{n} $ and we have : $$H_{n}=log(n)+γ+\frac{1}{2n}-\frac{1}{12n^2}+\frac{1}{120n^4} \hspace{0.5cm}.\hspace{.1cm}.\hspace{.1cm}. $$ Does any representation of a similar type exist for arbitrary positive $a$ ?

As Felix Marin answered,$$\sum_{k = 1}^{n}{1 \over 1 + ak}=\frac{H_{n+\frac{1}{a}}-H_{\frac{1}{a}}}{a}$$ Now, using the asymptotics $$H_m=\gamma +\log \left({m}\right)+\frac{1}{2 m}-\frac{1}{12
   m^2}+\frac{1}{120 m^4}+O\left(\frac{1}{m^5}\right)$$ $$H_{n+\frac{1}{a}}=\gamma +\log \left({n+\frac{1}{a}}\right)+\frac{1}{2 (n+\frac{1}{a})}-\frac{1}{12
   (n+\frac{1}{a})^2}+\frac{1}{120(n+\frac{1}{a} )^4}+O\left(\frac{1}{n^5}\right)$$ Now, developing each term as Taylor series $$\log \left({n+\frac{1}{a}}\right)=\log \left({n}\right)+\frac{1}{a n}-\frac{1}{2 a^2 n^2}+\frac{1}{3 a^3
   n^3}-\frac{1}{4 a^4 n^4}+O\left(\frac{1}{n^5}\right)$$ $$\frac{1}{ (n+\frac{1}{a})}=\frac{1}{n}-\frac{1}{a n^2}+\frac{1}{a^2 n^3}-\frac{1}{a^3
   n^4}+O\left(\frac{1}{n^5}\right)$$ $$\frac{1}{ (n+\frac{1}{a})^2}=\frac{1}{n^2}-\frac{2}{a n^3}+\frac{3}{a^2
   n^4}+O\left(\frac{1}{n^5}\right)$$  $$\frac{1}{ (n+\frac{1}{a})^4}=\frac{1}{n^4}+O\left(\frac{1}{n^5}\right) $$and replacing, you should get 
$$\sum_{k = 1}^{n}{1 \over 1 + ak}=\frac{\gamma-H_{\frac{1}{a}} }{a}+\frac{\log \left({n}\right) }{a}+\frac{a+2}{2 a^2
  }\frac 1{n}-\frac{a^2+6 a+6}{12 a^3 }\frac 1{n^2}+\frac{a^2+3 a+2}{6 a^4 }\frac 1{n^3}+\frac{a^4-30 a^2-60
   a-30}{120 a^5 }\frac 1{n^4}+O\left(\frac{1}{n^5}\right)$$
Take care that, for $a=1$ $$\sum_{k = 1}^{n}{1 \over 1 + k}=H_{n+1}-1=\gamma -1+\log \left({n}\right)+\frac{3}{2 n}-\frac{13}{12
   n^2}+\frac{1}{n^3}-\frac{119}{120
   n^4}+O\left(\frac{1}{n^5}\right)$$ and not $H_n$ (for which you gave the correct expansion) because of the shift of the index.
