Consider the series $$ \sum_{n=0}^{\infty}\left ( \frac{1}{n+1}-\frac{1}{z+n} \right )\tag{1} $$ It converges for all $z\notin \{0\}\cup \mathbb{Z}^-$. Does it imply, that $$ \sum_{n=0}^{\infty}\frac{1}{n+1} - \sum_{n=0}^{\infty}\frac{1}{z+n} $$ converges on that domain, even though each series diverges? Let us look at one example. Inserting $z=1/2$, the sum $(1)$ will be $-\log 4$, according to WolframAlpha. But is it true that $$ -\log 4=\sum_{n=0}^{\infty}\frac{1}{n+1} - \sum_{n=0}^{\infty}\frac{1}{\frac{1}{2}+n}? $$ I can not check it the right side on WolframAlpha or Maple. I recall that if two series converge, then combining it into one sum also converges. But I do not know if it holds conversely in general. Another similar question is, if two series are divergent, is the combined series divergent too? (I would answer no, not necessarily).

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    $\begingroup$ What if I do $\sum (n - n)$? Surely $\sum n$ doesn't converge. $\endgroup$
    – IAmNoOne
    Jun 21 '16 at 1:23
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    $\begingroup$ @Nameless Well, it's true. So there is only one way to talk about convergence, that is, if two series are convergent, then the combined series is convergent too, but not conversely, and it will never happen? $\endgroup$
    – Hopeless
    Jun 21 '16 at 1:29
  • $\begingroup$ "I can not check it the right side on WolframAlpha or Maple." Wait, you "cannot check" whether the series $\sum\limits_{n=0}^{\infty}\frac{1}{n+1}$ and $\sum\limits_{n=0}^{\infty}\frac{1}{\frac{1}{2}+n}$ converge or diverge? $\endgroup$
    – Did
    Jun 21 '16 at 4:52

If you are aware of harmonic numbers and digamma functions, then

$$S_1=\sum_{n=0}^{p} \frac{1}{n+1}=\psi (p+2)+\gamma=H_{p+1}$$ $$S_2=\sum_{n=0}^{p}\frac{1}{n+z}=\psi (p+z+1)-\psi(z) $$ Now, considering the asymptotics for large values of $p$ $$S_1-S_2=(\psi (z)+\gamma )+\frac{1-z}{p}+\frac{z^2+z-2}{2 p^2}+O\left(\frac{1}{p^3}\right)$$ which make $$\sum_{n=0}^{\infty}\left ( \frac{1}{n+1}-\frac{1}{z+n} \right )=\psi (z)+\gamma $$ You will find here some special values of the digamma function.


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