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: $$ \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}. $$
Explanation of Method
Since $(2n-1)^2\ge n^2$, each term of this series is bounded above by $1/n^2$, a convergent $p$-series ($\alpha=2>1$), so the comparison test immediately gives convergence. Going further, since $\sum 1/n^2$ splits exactly into its odd-indexed and even-indexed parts, and the even part is a scaled copy of the whole, algebra pins down the exact value of the odd part in terms of the well-known Basel sum $\pi^2/6$.
- Compare $\frac1{(2n-1)^2}$ against $\frac1{n^2}$ to establish convergence.
- Confirm via the limit comparison test.
- Split $\sum 1/n^2$ into odd and even parts to find the exact sum.
Step 1 · Direct comparison
For $n\ge1$, $2n-1\ge n$, so $(2n-1)^2\ge n^2$, giving:
$$ \frac{1}{(2n-1)^2} \le \frac{1}{n^2}. $$
Since $\sum_{n=1}^{\infty}\frac1{n^2}$ converges ($\alpha=2>1$, equal to $\pi^2/6$), by the comparison test $\sum_{n=1}^{\infty}\frac1{(2n-1)^2}$ converges too.
Step 2 · Confirm via the limit comparison test
With $b_n=\frac1{n^2}$:
$$ \lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\frac{n^2}{(2n-1)^2} = \lim_{n\to\infty}\frac{1}{4-\frac4n+\frac1{n^2}} = \frac14. $$
Positive and finite, so both series share the same convergence behavior — confirming convergence.
Step 3 · Find the exact sum by splitting odd and even terms
Splitting $\sum_{n=1}^{\infty}\frac1{n^2}$ into odd- and even-indexed terms:
$$ \sum_{n=1}^{\infty}\frac1{n^2} = \sum_{n=1}^{\infty}\frac1{(2n-1)^2} + \sum_{n=1}^{\infty}\frac1{(2n)^2} = \sum_{n=1}^{\infty}\frac1{(2n-1)^2} + \frac14\sum_{n=1}^{\infty}\frac1{n^2}. $$
Solving for the odd part:
$$ \frac34\sum_{n=1}^{\infty}\frac1{n^2} = \sum_{n=1}^{\infty}\frac1{(2n-1)^2} \qquad\Longrightarrow\qquad \sum_{n=1}^{\infty}\frac1{(2n-1)^2} = \frac34\cdot\frac{\pi^2}{6} = \frac{\pi^2}{8}. $$
Summary Table
| Quantity | Value |
|---|---|
| Comparison bound | $\dfrac1{(2n-1)^2}\le\dfrac1{n^2}$ |
| Limit comparison ratio | $\dfrac14$ |
| Sum $\sum 1/(2n-1)^2$ | $\dfrac{\pi^2}{8}$ |
Additional Notes
- This problem is the exponent-$2$ counterpart to Problem 27's exponent-$1$ case: raising the exponent from $1$ to $2$ flips the conclusion from divergence to convergence, illustrating the sharp threshold at $\alpha=1$.
- The technique of splitting $\zeta(2)=\pi^2/6$ into odd and even parts (via the even part being $\frac14\zeta(2)$) is a standard tool for evaluating sums restricted to arithmetic progressions of indices, and generalizes to other even exponents.
- As a byproduct, the even-index sum is also determined: $\sum_{n=1}^{\infty}\frac1{(2n)^2} = \frac14\cdot\frac{\pi^2}6 = \frac{\pi^2}{24}$, and indeed $\frac{\pi^2}{8}+\frac{\pi^2}{24} = \frac{3\pi^2+\pi^2}{24} = \frac{4\pi^2}{24} = \frac{\pi^2}6$, consistent with the original Basel sum.
Result
The series converges, with exact sum: