Prove that the series sum from n=1 to infinity 1/(a_n)
Prove that the series $\sum_{n=1}^{\infty} \frac{1}{a_n}$ diverges if $a_1, a_2, a_3, \ldots$ is an arithmetic sequence.
Explanation of Method
An arithmetic sequence $a_n=a_1+(n-1)d$ grows (at most) linearly in $n$, so its reciprocal behaves like a constant multiple of $\frac1n$. Since the harmonic series is the borderline divergent case, any sequence whose reciprocal is asymptotically proportional to $\frac1n$ — as every nonconstant arithmetic sequence is — inherits that divergence via the limit comparison test.
- Handle the degenerate case $d=0$ separately.
- Apply the limit comparison test against $\sum1/n$ for $d\neq0$.
- Confirm independently via the integral test.
Step 1 · The case $d=0$
If $d=0$, $a_n=a_1$ for every $n$, so $\sum\frac1{a_n}=\sum\frac1{a_1}=\infty$ trivially diverges.
Step 2 · Limit comparison test for $d\neq0$
Assume $d>0$ (a positive, eventually-positive arithmetic sequence). With $b_n=\dfrac1n$:
$$ \lim_{n\to\infty} \frac{1/a_n}{1/n} = \lim_{n\to\infty}\frac{n}{a_1+(n-1)d} = \lim_{n\to\infty}\frac{1}{\frac{a_1}n+d-\frac dn} = \frac1d. $$
Positive and finite; since $\sum\frac1n$ diverges, by the limit comparison test $\sum\frac1{a_n}$ diverges too.
Step 3 · Confirm via the integral test
Comparing with $\int_1^N \dfrac{dx}{a_1+(x-1)d}$, substituting $u=a_1+(x-1)d$:
$$ \int_1^N \frac{dx}{a_1+(x-1)d} = \frac1d\ln\frac{a_1+(N-1)d}{a_1} \to \infty \quad\text{as } N\to\infty. $$
The integral diverges, confirming that $\sum\frac1{a_n}$ diverges by the integral test as well.
Summary Table
| Case | Result |
|---|---|
| $d=0$ | $\sum\frac1{a_1}=\infty$; diverges trivially |
| $d\neq0$ | $\frac1{a_n}\sim\frac1{dn}$; diverges by limit/integral comparison |
Additional Notes
- Problem 27's odd-harmonic series $\sum\frac1{2n-1}$ is exactly the special case $a_1=1,d=2$ of this general result, so its earlier divergence proof is a direct corollary of this theorem.
- The result holds for any nonzero common difference, positive or negative (as long as the terms are eventually positive), since only the magnitude $|d|$ enters the leading-order asymptotic $\frac1{a_n}\sim\frac1{|d|n}$.
- This theorem is a clean structural statement: no matter how the arithmetic sequence is "tuned" (choice of $a_1$ and $d$), reciprocal summation always inherits the harmonic series' divergence — arithmetic growth is simply never fast enough to produce a convergent reciprocal series.
Result
For any arithmetic sequence $a_n$, the series diverges: