Solution Found!
Showing that In 1734, Leonhard Euler informally proved
Chapter 10, Problem 58AE(choose chapter or problem)
Showing that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}\) In 1734, Leonhard Euler informally proved that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}\). An elegant proof is outlined here that uses the inequality.
\(\cot ^{2} x<\frac{1}{x^{2}}<1+\cot ^{2} x\) (provided that \(0<x<\pi / 2\)) and the identity
\(\sum_{k=1}^{n} \cot ^{2}(k \theta)=\frac{n(2 n-1)}{3}\), for n = 1, 2, 3 . . ., where \(\theta=\frac{\pi}{2 n+1}\).
a. Show that \(\sum_{k=1}^{n} \cot ^{2}(k \theta)<\frac{1}{\theta^{2}} \sum_{k=1}^{n} \frac{1}{k^{2}}<n+\sum_{k=1}^{n} \cot ^{2}(k \theta)\).
b. Use the inequality in part (a) to show that
\(\frac{n(2 n-1) \pi^{2}}{3(2 n+1)^{2}}<\sum_{k=1}^{n} \frac{1}{k^{2}}<\frac{n(2 n+2) \pi^{2}}{3(2 n+1)^{2}}\).
c. Use the Squeeze Theorem to conclude that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}\).
[Source: The College Mathematics Journal, 24, No. 5 (November, 1993).]
Questions & Answers
QUESTION:
Showing that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}\) In 1734, Leonhard Euler informally proved that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}\). An elegant proof is outlined here that uses the inequality.
\(\cot ^{2} x<\frac{1}{x^{2}}<1+\cot ^{2} x\) (provided that \(0<x<\pi / 2\)) and the identity
\(\sum_{k=1}^{n} \cot ^{2}(k \theta)=\frac{n(2 n-1)}{3}\), for n = 1, 2, 3 . . ., where \(\theta=\frac{\pi}{2 n+1}\).
a. Show that \(\sum_{k=1}^{n} \cot ^{2}(k \theta)<\frac{1}{\theta^{2}} \sum_{k=1}^{n} \frac{1}{k^{2}}<n+\sum_{k=1}^{n} \cot ^{2}(k \theta)\).
b. Use the inequality in part (a) to show that
\(\frac{n(2 n-1) \pi^{2}}{3(2 n+1)^{2}}<\sum_{k=1}^{n} \frac{1}{k^{2}}<\frac{n(2 n+2) \pi^{2}}{3(2 n+1)^{2}}\).
c. Use the Squeeze Theorem to conclude that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}\).
[Source: The College Mathematics Journal, 24, No. 5 (November, 1993).]
ANSWER:SOLUTION
Step 1
Given the inequality ,
and is also proved that