(i)$$\sum_{n=-\infty}^{\infty} \frac{1}{n+\alpha}$$ and (ii)$$\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{n+\alpha}$$

The double infinite sum is taken to be the Cauchy principle value,

$$\sum_{n=-\infty}^{\infty} = \lim_{N\to\infty} \sum_{n=-N}^{N}$$

Indeed, we have to consider the Cauchy principle value because (i) does not even convergence in the traditional sense, it is harmonic after all, only (ii) will convergence by Leibniz's alternating series test.

These are interesting sums, they sum to trigonometric functions.

Notice that $$\sum_{n=-N}^{N}\frac{1}{n+\alpha} = \frac{1}{z} + \sum_{n=1}^n \left( \frac{1}{\alpha + n} + \frac{1}{\alpha - n} \right)$$

Thus, (i) is equal to $$\frac{1}{\alpha}+\sum_{n=1}^{\infty} \frac{2\alpha}{\alpha^2 - n^2}$$.

Similarly, (ii) is equal to $$\frac{1}{\alpha}+\sum_{n=1}^{\infty} \frac{(-1)^n2\alpha}{\alpha^2 - n^2}$$.

Now we need a result from complex analysis which can be found in any standard textbook on complex analysis.

**Theorem:**Let f be a meromorphic function with finitely many poles at $$\omega_1,...,\omega_k$$ none of which are integers. Define g to be $$\pi f(z) \cot \pi z$$ and h to be $$\pi f(z) \csc \pi z$$. Let $$g_j$$ be the reside of g at $$\omega_j$$ and $$h_j$$ be the reside of h at $$\omega_j$$. Assume that $$f(z)=O(z^{-2})$$. Then,

$$\sum_{n=-\infty}^{\infty} f(n) = -\sum_{j=1}^k g_j$$ and $$\sum_{n=-\infty}^{\infty}(-1)^n f(n) = - \sum_{j=1}^k h_j$$

In our case we will let $$f(z) = \frac{\alpha}{\alpha^2 - z^2}$$. Clearly, $$f(z) = O(z^{-2})$$, that is, there is R>0 large enough so that $$|f(z)| \leq A|z|^{-2}$$ for some constant A. We also see that f(z) is meromorphic with the only poles at $$\pm \alpha$$ none of which are integers. So the theorem will apply.

Notice that,

$$\sum_{n=-\infty}^{\infty} \frac{\alpha}{\alpha^2 - n^2} = \frac{1}{\alpha} + \sum_{n=-\infty}^1 \frac{\alpha}{\alpha^2 -n^2} + \sum_{n=1}^{\infty}\frac{\alpha}{\alpha^2 - n^2}$$

This simplies to,

$$\frac{1}{\alpha}+\sum_{n=-\infty}^{\infty} \frac{2\alpha}{\alpha^2 - n^2} = \sum_{n=-\infty}^{\infty} \frac{1}{n+\alpha}$$

Thus, we see that,

$$\sum_{n=-\infty}^{\infty}\frac{1}{n+\alpha} = \sum_{n=-\infty}^{\infty} f(n)$$

Similarly,

$$\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{n+\alpha} = \sum_{n=-\infty}^{\infty} (-1)^n f(n)$$

It is easy to compute, $$g_1=g_2=-\tfrac{\pi}{2}\cot \pi \alpha$$ and $$h_1=h_2 = -\tfrac{\pi}{2}\csc \pi \alpha$$.

Thus, we have derived that:

1)$$\sum_{n=-\infty}^{\infty} \frac{1}{n+\alpha} = \pi \cot \pi \alpha, ~ ~ \alpha \not \in \mathbb{Z}$$

2)$$\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n+\alpha} = \pi \csc \pi \alpha, ~ ~ \alpha \not \in \mathbb{Z}$$

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