At the moment, I am investigating the Riesz means defined as the series $$ s^{\delta}(\lambda)=\sum_{n\leq\lambda}\left( 1-\frac{n}{\lambda} \right)^{\delta}s_n. $$ Consider the special case $s_n=1$. Then we have the representation $$ \sum_{n\leq\lambda}\left( 1-\frac{n}{\lambda} \right)^{\delta}=\frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}\frac{\Gamma(1+\delta)\Gamma(s)}{\Gamma(1+\delta+s)}\zeta(s)\lambda^sds=\frac{\lambda}{1+\delta}+\sum_n b_n\lambda^{-n} $$ (according to Wikipedia) where $\Gamma$ is the gamma function $$ \Gamma(t)=\int^{\infty}_{0}x^{t-1}e^{-x}dx $$ and $\zeta$ is the Riemann zeta function $$ \zeta(s)=\sum^{\infty}_{n=1}\frac{1}{n^s}. $$ A recommendation to solve this integral is to use Perron's formula $$ A(x)\sum_{n\leq x}a(n)=\frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}g(z)\frac{x^z}{z}dz $$ where we have the uniformly convergent Dirichlet series $$ g(s)=\sum^{\infty}_{n=1}\frac{a(n)}{n^s},~\Re(s)>\sigma,~s\in\mathbb{C} $$ for an arithmetic function $a(n)$. Applying the Perron formula to the Riesz means gives $$ \sum_{n\leq \lambda}\left( 1-\frac{n}{\lambda} \right)^{\delta}=\frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}\sum^{\infty}_{n=1}\frac{\left( 1-\frac{n}{\lambda} \right)^{\delta}}{n^s}\frac{x^z}{z}dz. $$ The key next step appears to be evaluating the infinite sum, but I am not sure how to do it. Any ideas?
Thanks in advance for your comments and help.