TITLE: double summation with square root
QUESTION [0 upvotes]: This is below-average question on this platform, but I am confused over this double summation with a square root:
$$\sum_{p=1}^2\sum_{q=1}^2\sqrt[2]{(p-.5)^2+(q-.5)^2}$$
Does it give an answer of $\sqrt{10}$ or $5.9907$?

REPLY [2 votes]: As said by vondbrand, just develop the summation. Start with the inner sum; so you have first $$\sum_{p=1}^2\sum_{q=1}^2\sqrt[2]{(p-\frac{1}{2})^2+(q-\frac{1}{2})^2}=\sum _{p=1}^2  \left(\sqrt{\left(p-\frac{1}{2}\right)^2+\frac{1}{4}}+\sqrt{\left(p-\frac{1}{2}\right)^2+\frac{9}{4}}\right)$$
Then do the same for the outer sum and you should arrive to   $$2 \sqrt{2}+\sqrt{10}$$ which is almost $5.99017$
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