Let $f\in S_k(\mathrm{SL}_2(\mathbb{Z}))$ be a cusp eigenfunction of all Hecke operators $T_nf=\lambda(n)f,$ then $f$ has the following Fourier expansion at the cusp $\infty$: $$f(z)=\sum_{n\ge 1}a(n)q^n\quad q:=e^{2\pi inz}$$ suppose the $f$ normalized ( $a(1) = 1.$) Then we have $$\lambda(n)=\frac{a(n)}{n^{\frac{k-1}{2}}}.$$ Can someone help me to prove the following asymptotic formula : $$\sum_{n\le x}\lambda(n)^2=c.x+O(x^{3/5})\quad (*)$$
I think that we must start with $$\sum_{n\le x}\lambda(n)^2=\sum_{n\le x}\frac{a(n)^2}{n^{k-1}}=\frac{A(x)}{x^{k-1}}+(k-1)\int_1^x\frac{A(t)}{t^k}dt$$ where $A(x):=\sum_{n\le x}a(n)^2$ and then use Rankin's paper to get $(*)$
I just have accessed to Rankin's paper in which he proves that if $f(z)=\sum_{n\ge1}a(n)q^n\in S_{k}(N,\chi)$ is a modular form we have $$A(x):=\sum_{n\le x}|a(n)|^2=\alpha x^k+O(x^{k-2/5})$$ If we assume in additional that $f$ is an eigenfunction of all Hecke operators $T_nf=λ_f(n)f$ and $a(1)=1$ then $$\lambda_f(n)=\frac{a(n)}{n^{(k-1)/2}}$$ By partial summation we have \begin{align} \sum_{n\le x}|\lambda_f(n)|^2 & = \sum_{n\le x}\frac{|a(n)|^2}{n^{k-1}}\\ & = \frac{A(x)}{x^{k-1}}+(k-1)\int_1^x\frac{A(t)}{t^k}dt \\ & =\alpha.x+O(x^{3/5})+(k-1)\int_1^x(\alpha+O(t^{-2/5}))dt\\ &=c.x+O(x^{3/5}) \end{align} where $c=\alpha.k$