Investigate whether the sum (or difference) of two
Investigate whether the sum (or difference) of two divergent series can itself converge. That is: if $\sum a_n$ and $\sum b_n$ both diverge, can $\sum(a_n+b_n)$ or $\sum(a_n-b_n)$ converge?
Explanation of Method
This is a theorem/counterexample problem, not a numeric convergence test. The standard theorem states that if $\sum a_n$ converges and $\sum b_n$ diverges, then $\sum(a_n\pm b_n)$ must diverge — but this says nothing about the case where both diverge. A concrete counterexample using two divergent harmonic-type series with a telescoping difference resolves the question decisively: yes, the difference of two divergent series can converge, while their sum, in the same example, remains divergent.
- Recall why "convergent $\pm$ divergent = divergent" always holds (for contrast).
- Construct an explicit pair of divergent series whose difference converges.
- Verify that the corresponding sum, for the same pair, still diverges.
Step 1 · The safe case: convergent + divergent
If $\sum a_n$ converges and $\sum b_n$ diverges, then $\sum(a_n+b_n)$ must diverge: if it converged, then $b_n=(a_n+b_n)-a_n$ would be a difference of two convergent series, hence $\sum b_n$ would converge — a contradiction. This argument fails entirely when both $\sum a_n,\sum b_n$ diverge, since there is no convergent series to isolate $b_n$ against.
Step 2 · A counterexample: the difference converges
Let $a_n=\dfrac1n$ and $b_n=\dfrac1{n+1}$. Both $\sum a_n=\sum\frac1n$ and $\sum b_n=\sum\frac1{n+1}$ diverge (both are harmonic-type series, differing only by an index shift). But:
$$ \sum_{n=1}^{\infty}(a_n-b_n) = \sum_{n=1}^{\infty}\left(\frac1n-\frac1{n+1}\right) = 1 $$
by direct telescoping: the $N$th partial sum is $1-\dfrac1{N+1}\to1$. So the difference of two divergent series converges here.
Step 3 · The sum, for the same pair, still diverges
$$ \sum_{n=1}^{\infty}(a_n+b_n) = \sum_{n=1}^{\infty}\left(\frac1n+\frac1{n+1}\right) \ge \sum_{n=1}^{\infty}\frac2{n+1}, $$
which diverges by comparison with the (scaled) harmonic series. So in this same example, the sum remains divergent while the difference converges — confirming that no general rule links the convergence of $a_n\pm b_n$ to that of $a_n,b_n$ individually once both are divergent.
Summary Table
| Series | Behavior |
|---|---|
| $\sum1/n$ | Diverges |
| $\sum1/(n+1)$ | Diverges |
| $\sum\left(\frac1n-\frac1{n+1}\right)$ | Converges (to $1$) |
| $\sum\left(\frac1n+\frac1{n+1}\right)$ | Diverges |
Additional Notes
- The mechanism here is exactly the telescoping cancellation seen throughout the telescoping-series problems earlier in this sequence: the leading-order divergence of $1/n$ and $1/n+1$ cancels almost entirely, leaving only a convergent "boundary" contribution, while the sum has no such cancellation and both divergences add constructively.
- This example generalizes immediately: for any divergent series $\sum c_n$ with $c_n\to0$, setting $a_n=c_n$ and $b_n=c_{n+1}$ (or any similar shift) produces a telescoping difference $\sum(a_n-b_n)=c_1-\lim c_n=c_1$, always convergent, regardless of how slowly or quickly $\sum c_n$ diverges.
- This problem is the natural companion to Problem 137, which asks the analogous question for the Cauchy product of two series rather than their sum/difference: both problems illustrate that formal algebraic operations on series do not preserve convergence/divergence properties in any simple, universal way — the specific structure of the terms always matters.
Result
Yes — the sum (or difference) of two divergent series can converge: