# How do i evaluate this :$\sum_{k=1}^{+\infty}\frac{{(-1)^k}}{\sqrt{k+1}+\sqrt{k}}$?

I have tried to evaluate the bellow sum using some standard altern sum but i don't succed then my question here is :

Question: How do i evaluate this sum

$\sum_{k=1}^{+\infty}\frac{{(-1)^k}}{\sqrt{k+1}+\sqrt{k}}$ ?

Note: Wolfram alpha show it's values here

• It doesn't converge. $\sum_{k=1}^{+\infty}\frac{{(-1)^k}}{\sqrt{k+1}+\sqrt{k}}$ does. – Professor Vector Nov 30 '17 at 21:03
• sorry i meant + in denominator – zeraoulia rafik Nov 30 '17 at 21:05
• That's better. But it's elementary that it converges, because it's an alternating series, and $\frac1{\sqrt{k+1}+\sqrt{k}}$ is monotone decreasing. Calculating the sum is another business. =D – Professor Vector Nov 30 '17 at 21:08
• Since $$\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k} = \int_{0}^{+\infty} \left(\frac{1-e^{-s}}{2s^{3/2}\sqrt{\pi}}\right)e^{-kx}\,dx \tag{A}$$ we have: $$\sum_{k\geq 1}\frac{(-1)^k}{\sqrt{k}+\sqrt{k+1}} = -\frac{1}{2\sqrt{\pi}}\int_{0}^{+\infty}\frac{1-e^{-s}}{s^{3/2}(e^s+1)}\,ds \tag{B}$$ and the series in the LHS is clearly convergent by Leibniz' test. – Jack D'Aurizio Nov 30 '17 at 21:17
• The RHS of (B) gives a relation between the values of the $\zeta$ function at half-integers and the LHS. – Jack D'Aurizio Nov 30 '17 at 21:19

Since $\frac1{\sqrt{k}+\sqrt{k+1}} =\sqrt{k+1}-\sqrt{k}$, the sum of consecutive odd and even terms is

$\begin{array}\\ -\frac{1}{\sqrt{2k}+\sqrt{2k-1}}+\frac{1}{\sqrt{2k+1}+\sqrt{2k}} &=-(\sqrt{2k}-\sqrt{2k-1})+(\sqrt{2k+1}-\sqrt{2k})\\ &=\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\\ &=\sqrt{2k}(\sqrt{1+(1/2k)}-2+\sqrt{1-1/(2k)})\\ &=\sqrt{2k}(1+\frac1{4k}-\frac1{32k^2}+O(1/k^3)-2+1-\frac1{4k}-\frac1{32k^2}+O(1/k^3))\\ &=\sqrt{2k}(-\frac1{16k^2}+O(1/k^3))\\ &=-\dfrac{\sqrt{2}}{16k^{3/2}}(1+O(1/k))\\ \end{array}$

Taking more terms in the expansion of $\sqrt{1+x}$ would give expressions involving $\frac1{k^{m+1/2}}$ for $m=2. 3. ...$.

Summing these involves $\zeta(m+1/2)$.