Is the series sum from n=1 to infinity 1/(a_n) convergent
Is the series $\sum_{n=1}^{\infty} \frac{1}{a_n}$ convergent if $a_1, a_2, a_3, \ldots$ is a geometric sequence?
Explanation of Method
A geometric sequence $a_n=a_1r^{n-1}$ has reciprocals $\frac1{a_n}=\frac1{a_1}\left(\frac1r\right)^{n-1}$ — themselves a geometric series, but with the reciprocal ratio $\frac1r$. Convergence of a geometric series requires its ratio to have absolute value less than $1$, so the reciprocal series converges exactly when $|1/r|<1$, i.e. $|r|>1$: the opposite condition to the original sequence's own convergence.
- Rewrite the reciprocal series as a geometric series with ratio $1/r$.
- Apply the geometric series convergence criterion.
- Check the boundary cases $|r|=1$ separately.
Step 1 · Rewrite as a geometric series
With $a_n=a_1r^{n-1}$ ($a_1>0$):
$$ \sum_{n=1}^{\infty}\frac1{a_n} = \frac1{a_1}\sum_{n=1}^{\infty}\left(\frac1r\right)^{n-1}. $$
Step 2 · Apply the geometric series criterion
This converges iff $\left|\frac1r\right|<1$, i.e. $|r|>1$, in which case:
$$ \sum_{n=1}^{\infty}\frac1{a_n} = \frac1{a_1}\cdot\frac1{1-\frac1r} = \frac{r}{a_1(r-1)}. $$
Step 3 · Check the boundary cases
If $r=1$: $a_n=a_1$ for all $n$, so $\sum\frac1{a_n}=\sum\frac1{a_1}=\infty$; diverges. If $r=-1$: $\frac1{a_n}$ alternates in sign without shrinking, and the series oscillates without converging. If $|r|<1$: the original sequence $a_n\to0$, so $\frac1{a_n}\to\infty$, and the reciprocal series (itself geometric with ratio $\frac1r$, $\left|\frac1r\right|>1$) diverges.
Summary Table
| Value of $r$ | Convergence |
|---|---|
| $|r|>1$ | Converges to $\dfrac{r}{a_1(r-1)}$ |
| $|r|=1$ | Diverges (or oscillates) |
| $|r|<1$ | Diverges |
Additional Notes
- This is the exact mirror image of Problem 53: arithmetic sequences always have divergent reciprocal series, while geometric sequences can go either way depending on whether $|r|$ exceeds $1$.
- The condition flips because reciprocating a geometric sequence just inverts its ratio ($r\mapsto1/r$), whereas reciprocating an arithmetic sequence produces a fundamentally different (non-arithmetic, non-geometric) sequence whose asymptotic behavior always resembles the harmonic series.
- Concretely: $a_n=2^n$ ($r=2>1$) gives the convergent series $\sum2^{-n}=1$; $a_n=(1/2)^n$ ($r=\frac12<1$) gives the divergent series $\sum2^n=\infty$ — exact reciprocal roles depending purely on whether the original sequence grows or decays.
Result
The convergence depends entirely on the common ratio $r$: