# sum of two series

Can $$\sum_{k= 0}^{\infty}\frac{\left( -1\right) ^{k}}{k^2+a^{2}}, \ \ \sum_{k= 0}^{\infty}\frac{\left( -1\right) ^{k}}{(k-1/2)^2+a^{2}}$$ be summed explicitly, where $a$ is a constant real number? Can any one give me some hint or tell me that the analytic expression doesn't exist. Thanks very much!

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$$\sum_{k=0}^\infty\frac{(-1)^k}{k^2+a^2}=\frac{1+a\pi\cdot\text{csch }(a\pi)}{2a^2}\qquad;\qquad\sum_{k=0}^\infty\frac{(-1)^k}{k^2-a^2}=-\frac{1+a\pi‌​\cdot\csc(a\pi)}{2a^2}$$ – Lucian Mar 26 '14 at 0:09
Maple can give closed forms for the sums. I have not tried Mathematics though. – Mhenni Benghorbal Mar 26 '14 at 22:49
@MhenniBenghorbal: they hopefully look like my answer. – robjohn Mar 26 '14 at 22:49
@robjohn: Maple gives the answers in terms of the digamma function as yours, however you can simplify further. – Mhenni Benghorbal Mar 26 '14 at 22:51
@MhenniBenghorbal: the answer to the second sum is in terms of the digamma function; the first sum is in terms of the hyperbolic cosecant function. – robjohn Mar 26 '14 at 22:53

We can use equation $(7)$ from this answer: $$\sum_{k\in\mathbb{Z}}\frac{1}{k+z}=\pi\cot(\pi z)\tag{1}$$ Directly from $(1)$, we have $$\sum_{k\in\mathbb{Z}}\frac2{2k+z}=\pi\cot(\pi z/2)\tag{2}$$ Subtracting $(1)$ from $(2)$ yields \begin{align} \pi\csc(\pi z) &=\sum_{k=-\infty}^\infty\frac{(-1)^k}{k+z}\tag{3}\\ &=\frac1z-2z\sum_{k=1}^\infty\frac{(-1)^k}{k^2-z^2}\tag{4} \end{align} Substituting $z\mapsto iz$ in $(4)$ gives $$\pi\,\mathrm{csch}(\pi z) =\frac1z+2z\sum_{k=1}^\infty\frac{(-1)^k}{k^2+z^2}\tag{5}$$ This works nicely to give us the first sum: $$\color{#C00000}{\sum_{k=0}^\infty\frac{(-1)^k}{k^2+z^2} =\frac{\pi z\,\mathrm{csch}(\pi z)+1}{2z^2}}\tag{6}$$

Substituting $z\mapsto z+\frac12$ in $(3)$ gives \begin{align} \pi\sec(\pi z) &=\sum_{k=-\infty}^\infty\frac{(-1)^k}{k+\frac12+z}\tag{7}\\ &=\sum_{k=0}^\infty\frac{(-1)^k(2k+1)}{\left(k+\frac12\right)^2-z^2}\tag{8} \end{align} Substituting $z\mapsto iz$ in $(8)$ gives $$\pi\,\mathrm{sech}(\pi z) =\sum_{k=0}^\infty\frac{(-1)^k(2k+1)}{\left(k+\frac12\right)^2+z^2}\tag{9}$$ This fails to give us the second sum because we have to add $\frac1{k+\frac12+z}$ and $\frac1{k+\frac12-z}$ instead of subtracting them because of the alternating signs. So we have to resort to another approach.

The digamma function $$\psi(x)=-\gamma+\sum_{k=0}^\infty\left(\frac1{k+1}-\frac1{k+x}\right)\tag{10}$$ is the derivative of the logarithm of the Gamma function. $$\psi\left(-\frac14+\frac{iz}{2}\right)-\psi\left(-\frac14-\frac{iz}{2}\right) =4iz\sum_{k=0}^\infty\frac1{\left(2k-\frac12\right)^2+z^2}\tag{11}$$ and $$\psi\left(\frac14+\frac{iz}{2}\right)-\psi\left(\frac14-\frac{iz}{2}\right) =4iz\sum_{k=0}^\infty\frac1{\left(2k+\frac12\right)^2+z^2}\tag{12}$$ Subtracting $(12)$ from $(11)$ yields the second sum: \begin{align} &\hspace{-1cm}\color{#C00000}{\sum_{k=0}^\infty\frac{(-1)^k}{\left(k-\frac12\right)^2+z^2}}\\[6pt] &\hspace{-1cm}\color{#C00000}{=\frac1{4iz}\left(\psi\left(-\frac14+\frac{iz}{2}\right)-\psi\left(-\frac14-\frac{iz}{2}\right)-\psi\left(\frac14+\frac{iz}{2}\right)+\psi\left(\frac14-\frac{iz}{2}\right)\right)}\tag{13} \end{align}

The second sum without alternation can be achieved by substituting $z\mapsto z-\frac12$ in $(1)$: \begin{align} -\pi\tan(\pi z) &=\sum_{k\in\mathbb{Z}}\frac1{k-\frac12+z}\tag{14}\\ &=-2z\sum_{k=1}^\infty\frac1{\left(k-\frac12\right)^2-z^2}\tag{15} \end{align} Substituting $z\mapsto iz$ in $(15)$ yields $$\pi\tanh(\pi z)=2z\sum_{k=1}^\infty\frac1{\left(k-\frac12\right)^2+z^2}\tag{16}$$ Thus, we get the second sum without alternation $$\sum_{k=0}^\infty\frac1{\left(k-\frac12\right)^2+z^2} =\frac4{4z^2+1}+\frac{\pi\tanh(\pi z)}{2z}\tag{17}$$ From $(12)$, we get $$\sum_{k=0}^\infty\frac2{\left(2k+\frac12\right)^2+z^2} =\frac1{2iz}\left(\psi\left(\frac14+\frac{iz}{2}\right)-\psi\left(\frac14-\frac{iz}{2}\right)\right)\tag{18}$$ Subtracting $(18)$ from $(17)$ yields an alternate form for the second sum: \begin{align} &\color{#C00000}{\sum_{k=0}^\infty\frac{(-1)^k}{\left(k-\frac12\right)^2+z^2}}\\[6pt] &\color{#C00000}{=\frac4{4z^2+1}+\frac{\pi\tanh(\pi z)}{2z}-\frac1{2iz}\left(\psi\left(\frac14+\frac{iz}{2}\right)-\psi\left(\frac14-\frac{iz}{2}\right)\right)}\tag{19} \end{align}

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It is well known that $$\csc z = \frac{1}{z} + 2z\sum_{k = 1}^\infty \frac{(-1)^k}{z^2 - k^2 \pi^2}.$$ Changing $z$ into $\pi z$ produces $$\pi\csc(\pi z) = \frac{1}{z} + 2z\sum_{k = 1}^\infty \frac{(-1)^k}{z^2 - k^2}, \quad z \ne n \pi, \quad n \in \mathbb{Z},$$ so $$\frac{\pi\csc(\pi z) - 1/z}{2z} = \sum_{k = 1}^\infty \frac{(-1)^k}{z^2 - k^2}.$$ Then it can be shown, by replacing $z$ with $iz$ and using $\csc(iz) = -i\operatorname{cosech} z$, that $$\operatorname{cosech} z = \frac{1}{z} + 2z\sum_{k = 1}^\infty \frac{(-1)^k}{z^2 + k^2}, \quad z \ne n \pi i, \quad n \in \mathbb{Z},$$ or $$\frac{\pi\operatorname{cosech}(\pi z) - 1/z}{2z} = \sum_{k = 1}^\infty \frac{(-1)^k}{z^2 + k^2}.$$

The second series looks familiar, but I cannot remember what it equals.

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The $k=0$ term is missing, but otherwise, this gives the correct answer for the first sum. – robjohn Mar 27 '14 at 0:15

$$\sum_{k= 0}^{\infty}\frac{\left( -1\right) ^{k}}{k^2+a^{2}}$$ This series converges to $$\dfrac{1}{2a^2}+\dfrac{\pi\mathrm{csch}(\pi a)}{2a}$$ Now, $$\sum_{k= 0}^{\infty}\frac{\left( -1\right) ^{k}}{(k-1/2)^2+a^{2}}$$ This series also converges by the comparison test.

Now, $$\sum_{k= 0}^{\infty}\left(\frac{\left( -1\right) ^{k}}{k^2+a^{2}}+ \frac{\left( -1\right) ^{k}}{(k-1/2)^2+a^{2}}\right)$$ Use the comparison test to see that this is convergent.

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I think the asker wanted to find explicitely the sum of each series (individually), rather than just show that they are convergent. I could have misunderstood though – Hayden Mar 26 '14 at 0:01
@Hayden Maybe. I will edit my answer. – Sanath K. Devalapurkar Mar 26 '14 at 0:02
@Hayden See my edited answer. – Sanath K. Devalapurkar Mar 26 '14 at 0:07
Are you sure about the divergence of the sum? Wolfram alpha says it converges by the comparison test. – David H Mar 26 '14 at 0:42
@DavidH My mistake - I wrote it in a hurry. – Sanath K. Devalapurkar Mar 26 '14 at 0:48