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Let partial sums $s_{a,d,n}$ for positive integers $a,d,n$ be defined as $$s_{a,d,n}=\sum_{k=1}^{n}\frac{1}{a+(k-1)d}$$

with $$S_{a,d}=\lim_{n\to \infty}s_{a,d,n}$$

e.g. $$S_{1,1}=S=\frac 1 1+\frac12+\frac 13+...$$ and $$S_{1,2}=\frac 1 1+\frac13+\frac 15+...$$ By rearrnging terms in S, it can be shown that $${S_{1,2}\over S}=\frac12$$

  1. What is $S_{a,d}/S$ ? Does it make sense to take the ratio of infinite quantities? The motivation behind this question is: "if $1/1+1/2+1/3...$ is $S$ then proprtional to that what would be $S_{a,d}".$
  2. Is 1. equal to $$\lim_{n\to\infty}\frac{s_{a,d,n}}{s_{1,1,n}}$$

Working it out, it seems that , $$\lim_{n\to\infty}\frac{s_{a,d,n}}{s_{1,1,n}}=\lim_{n\to\infty} \frac1d\frac{\psi(n+a/d)-\psi(a/d)}{\psi(n+a)-\psi(a)}$$ where $\psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}$ is digamma.
3. Is the RHS $\frac 1d$? It seems so heurestically but i am not sure how to go about proving that.

Jam
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lineage
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  • It only makes sense to take the ratio of divergent series if they are treated with a particular regularization method, which you seem to have missed out. In terms of normal limits, $S$ is undefined, so you reach an impasse. Or alternatively, you could take the ratio of the partial series, which may be well defined. The answer to your second question is contingent on the treatment in the first. – Jam Dec 13 '19 at 17:59
  • @Jam whats regularization?Assume S is regularized. – lineage Dec 13 '19 at 18:01
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    Since we can show that $S$ doesn't convergent to any value (except possibly $\infty$), it doesn't make sense to talk about it. Unless, you've broadened the definition of how you assign values to divergent series. Regularization is the methods for assigning values to divergent series. See Question 1085570, Wikipedia: Divergent Series and Wikipedia: Regularization (Physics). – Jam Dec 13 '19 at 18:06
  • There isn't just one single way of assigning a value to divergent series, and they don't all work on all cases. A common type of regularization called Zeta function regularization would still not work on $S$. So your question is incomplete, without this information. – Jam Dec 13 '19 at 18:11
  • thnx for the help...does 3. make sense to ask ? – lineage Dec 13 '19 at 18:16
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    Yes and the RHS is indeed $\frac1d$, since $\psi(n+x)$ dominates $\psi(x)$ as $n\to\infty$, so $\frac{\psi(n+a/d)-\psi(a/d)}{\psi(n+a)-\psi(a)}\to\frac{\psi(n+a/d)}{\psi(n+a)}\to 1$. – Jam Dec 13 '19 at 18:25

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