Knowing that sum from n=1 to infinity 1/(n^()alpha)
Knowing that $\sum_{n=1}^{\infty}\frac{1}{n^\alpha}$ converges for $\alpha>1$ and diverges for $\alpha\le1$, determine the convergence of: $$ \sum_{n=1}^{\infty} \frac{1}{2n-1}. $$
Explanation of Method
The terms $\frac1{2n-1}$ are the reciprocals of odd numbers, and they are comparable in size to $\frac1{2n}$ — half the harmonic series terms. Since the harmonic series $\sum\frac1n$ diverges ($\alpha=1$), and $\frac1{2n-1}\ge\frac1{2n}$ termwise, the comparison test (or, equivalently, a limit-comparison / algebraic contradiction argument) forces this odd-harmonic series to diverge as well.
- Compare $\frac1{2n-1}$ termwise against $\frac1{2n}$.
- Apply the comparison test using the known divergence of the harmonic series.
- Confirm via the limit comparison test.
Step 1 · Direct comparison
For $n\ge1$, $2n-1\le2n$, so:
$$ \frac{1}{2n-1} \ge \frac{1}{2n}. $$
Summing both sides:
$$ \sum_{n=1}^{\infty}\frac{1}{2n-1} \ge \sum_{n=1}^{\infty}\frac1{2n} = \frac12\sum_{n=1}^{\infty}\frac1n. $$
Step 2 · Apply the comparison test
Since $\alpha=1$, the series $\sum\frac1n$ diverges, so $\frac12\sum\frac1n$ diverges too. By the comparison test, since $\frac1{2n-1}\ge\frac1{2n}$ and $\sum\frac1{2n}$ diverges, the series $\sum_{n=1}^{\infty}\frac1{2n-1}$ diverges as well.
Step 3 · Confirm via the limit comparison test
With $b_n=\frac1n$:
$$ \lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\frac{n}{2n-1} = \lim_{n\to\infty}\frac{1}{2-\frac1n} = \frac12. $$
Since this limit is positive and finite, and $\sum\frac1n$ diverges, $\sum\frac1{2n-1}$ diverges as well — consistent with the direct comparison.
Summary Table
| Method | Result |
|---|---|
| Direct comparison | $\dfrac1{2n-1}\ge\dfrac1{2n}$, and $\sum\dfrac1{2n}$ diverges |
| Limit comparison with $1/n$ | ratio $\to\dfrac12>0$ |
| Conclusion | Diverges |
Additional Notes
- An algebraic shortcut: if $\sum\frac1{2n-1}$ converged to some $S$, then since $\sum\frac1n = \sum\frac1{2n-1}+\sum\frac1{2n} = S+\frac12\sum\frac1n$, solving gives $\sum\frac1n = 2S$ — a finite value, contradicting the known divergence of the harmonic series.
- The odd-harmonic series grows asymptotically like $\frac12\ln n$, exactly half the rate of the full harmonic series ($\sim\ln n$), since it comprises roughly half of the harmonic terms with comparable magnitude.
- This is a clean illustration of why the exponent $\alpha=1$ is the critical threshold: any series whose terms decay like a constant times $1/n$ — whether the harmonic series itself or a comparable variant like this one — diverges.
Result
The series diverges: