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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?

user190080
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soodehMehboodi
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  • It might be slightly nicer to rewrite it to $$ a_n = \frac12\sum_{k=1}^n \left(3\sqrt k - \Bigl(2\sqrt{k+1/2}+\sqrt{k-1}\Bigr)\right) $$ Then, intuitively, the terms go asymptotically like the second derivative of $\sqrt k$. – hmakholm left over Monica Apr 13 '16 at 18:02

1 Answers1

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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:

$$\frac{(\sqrt2-4) \zeta(3/2)+4 \pi\sqrt 2}{8 \pi}$$

Thomas Andrews
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