Is it true that $$\sum_{k=1}^\infty \left(\frac 3 2\sqrt{k}-\sqrt{k+1/2}-\frac{1}{2}\sqrt{k-1}\right)=\frac{(\sqrt2-4) \zeta(3/2)+4 \pi\sqrt 2}{8 \pi}?$$
(this is in regards to the question: Convergence of a compound sequence)
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It is enough to compute asymptotics for $$ S_n = \sum_{k=1}^{n}\sqrt{k} = \frac{2}{3}\,n^{3/2}+\frac{1}{2} n^{1/2}\color{red}{-\frac{\zeta\left(\frac{3}{2}\right)}{4\pi}}+O\left(\frac{1}{\sqrt{n}}\right)\tag{1}$$ and $$ T_n = \sum_{k=0}^{n}\sqrt{k+\frac{1}{2}} = \frac{1}{\sqrt{2}}\left(S_{2n+1}-\sqrt{2}\,S_n\right)\tag{2} $$ by summation by parts to find the value of the given series through straightforward algebraic manipulations. The relevant part is just the red term in $(1)$, that equals $\zeta\left(-\frac{1}{2}\right)$ by the reflection formula for the $\zeta$ function.
Jack D'Aurizio
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Put $$S_n=\sum_{k=1}^n (\frac{3}{2\sqrt{2}}\sqrt{2k}-\frac{1}{\sqrt{2}}\sqrt{2k+1}-\frac{1}{\sqrt{2}}\sqrt{2k-2})$$
and $\displaystyle T_n=\sum_{k=1}^n\sqrt{j}$. Using that $\displaystyle T_{2n+2}=\sqrt{2}T_n+\sqrt{2}\sqrt{n+1}+\sum_{j=1}^n \sqrt{2j+1}$, you get a formula for $S_n$ using $T_n$ and $T_{2n+2}$. Now using this answer Euler-Maclaurin Summation , you get asymptotic formula for $T_n$, and you can finish the job (with some computations..).
