Can Leibniz's test be applied to the series
Can Leibniz's test be applied to the series $$ \frac{1}{\sqrt{2}-1}-\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}+\cdots+\frac{1}{\sqrt{n}-1}-\frac{1}{\sqrt{n}+1}+\cdots\,? $$ Does the given series converge?
Explanation of Method
Although the signs visually alternate, they don't alternate term-by-term in the sense Leibniz's test requires — the plus/minus pattern instead groups terms in pairs, each pair belonging to the same index $n$. Combining each pair via a common denominator collapses it to a single positive term $\frac2{n-1}$, revealing that the "series" is secretly just (twice) the divergent harmonic series in disguise.
- Group the series by pairs and simplify each pair via a common denominator.
- Recognize that all resulting terms are positive, so Leibniz's test does not apply.
- Determine convergence by identifying the simplified series as a multiple of the harmonic series.
Step 1 · Simplify each pair
The series is $\displaystyle\sum_{n=2}^{\infty}\left(\frac1{\sqrt n-1}-\frac1{\sqrt n+1}\right)$. Combining over a common denominator:
$$ \frac1{\sqrt n-1}-\frac1{\sqrt n+1} = \frac{(\sqrt n+1)-(\sqrt n-1)}{(\sqrt n-1)(\sqrt n+1)} = \frac2{n-1}. $$
Step 2 · Leibniz's test does not apply
The simplified general term $a_n=\dfrac2{n-1}$ is positive for every $n\ge2$ — there is no alternating sign pattern in the actual sequence of terms. The visual $+/-$ signs in the original expression merely group two same-signed contributions within each pair; Leibniz's test, which requires strictly alternating signs, cannot be applied.
Step 3 · Determine convergence
Since $a_n=\dfrac2{n-1}$, the series is:
$$ \sum_{n=2}^{\infty}\frac2{n-1} = 2\sum_{k=1}^{\infty}\frac1k, $$
twice the harmonic series, which diverges. So the given series diverges.
Summary Table
| Quantity | Value |
|---|---|
| Simplified term | $\dfrac2{n-1}$ (always positive) |
| Leibniz's test applicable? | No — no genuine alternation |
| Conclusion | Diverges (twice the harmonic series) |
Additional Notes
- This problem is a deliberate warning against pattern-matching on visual sign sequences: the presence of $+$ and $-$ symbols in a written-out series is not sufficient grounds to invoke Leibniz's test — one must verify that consecutive terms of the sequence being summed actually alternate in sign.
- The rationalization technique used here (multiplying out $\frac1{\sqrt n-1}-\frac1{\sqrt n+1}$ over a common denominator) is the same conjugate-style simplification seen in Problem 99's $\sqrt{n+a}-\sqrt{n+b}$, again converting an awkward radical expression into a clean rational one.
- Structurally, this differs sharply from Problem 111, where a legitimate rearrangement of a genuinely alternating (and conditionally convergent) series produces a different but still finite sum — here, there was never true alternation to begin with, so no rearrangement subtlety is even relevant.
Result
Leibniz's test cannot be applied (no genuine sign alternation); the series diverges: