Convergence of a compound sequence let $a_n=\frac{1}{2}\sqrt{n}+\sum_{k=1}^n(\sqrt{k}-\sqrt{k+\frac{1}{2}})$ be a sequence.
Is this sequence convergent?

Write $\sqrt{n} = \sum_{k=1}^{n} \left(\sqrt{k}-\sqrt{k-1}\right)$. Then you get:
$$a_n = \sum_{k=1}^{n}\left(\sqrt{k}-\sqrt{k+1/2}+\frac{1}{2}\sqrt{k}-\frac{1}{2}\sqrt{k-1}\right)=\sum_{k=1}^n \left(\frac 3 2\sqrt{k}-\sqrt{k+1/2}-\frac{1}{2}\sqrt{k-1}\right)$$
This represents $a_n$ as a pure series, with $a_n=\sum_{k=1}^{n} b_k$ with $b_k=
\frac 3 2\sqrt{k}-\sqrt{k+1/2}-\frac{1}{2}\sqrt{k-1}$.
Now $$\sqrt{k+1/2}=\sqrt{k}{\sqrt{1+1/(2k)}} = \sqrt{k}+\frac{1}{4\sqrt{k}} + O\left(\frac{1}{k^{3/2}}\right)$$ and similarly $$\sqrt{k-1}=\sqrt{k}-\frac{1}{2\sqrt{k}}+O\left(\frac{1}{k^{3/2}}\right)$$ 
So you get
$$b_k=\frac 3 2\sqrt{k}-\sqrt{k+1/2}-\frac{1}{2}\sqrt{k-1}=O\left(\frac1{k^{3/2}}\right)$$
So $\sum_{k=1}^{\infty} b_k$ converges, and you are done.
Wolfram alpha gives an exact value of:
$$\frac{(\sqrt2-4) \zeta(3/2)+4 \pi\sqrt 2}{8 \pi}$$
