Sum to infinity of reciprocal of arithmetic sequence terms

Question

Is there a arithmetic sequence starting at any a, such that the sum of the reciprocals of its terms to infinity converges? i.e. $\sum_{n=0}^{\infty} \frac 1{a+nd}=\frac 1a+\frac 1{a+d}+\frac 1{a+2d} +\frac 1{a+3d}\cdots $ converges for any a or d

Answer 1

Simply observe that

$$\frac 1{a+nd}\sim \frac 1{nd}$$

then the given series diverges by limit comparison test with the harmonic series $\sum \frac 1{n}$.

Answer 2

Let the $n^{th}$ term be $a_n=\frac{1}{a+nd}$
Then, $a_n-a_{n-1}=\frac{-d}{a^2+n\left(n-1\right)d^2+\left(2n-1\right)}$
Which decreases at a very small rate which means the series diverges