Sum of 1/n2 or 61/36: Which is Bigger?

  • Context: MHB 
  • Thread starter Thread starter Albert1
  • Start date Start date
Click For Summary
SUMMARY

The comparison between the series $\sum \dfrac{1}{n^2}$ and the fraction $\dfrac{61}{36}$ reveals that $\sum \dfrac{1}{n^2}$ converges to $\dfrac{\pi^2}{6}$, which is approximately 1.64493. In contrast, $\dfrac{61}{36}$ equals approximately 1.69444. Therefore, $\dfrac{61}{36}$ is definitively larger than $\sum \dfrac{1}{n^2}$. This conclusion holds true regardless of the assumption that $\sum \dfrac{1}{n^2} = \dfrac{\pi^2}{6}$ is unknown.

PREREQUISITES
  • Understanding of infinite series and convergence
  • Familiarity with the Basel problem and its solution
  • Basic knowledge of mathematical constants, specifically $\pi$
  • Ability to perform numerical comparisons of rational and irrational numbers
NEXT STEPS
  • Study the derivation of the Basel problem and its historical significance
  • Explore the properties of convergent series in calculus
  • Learn about numerical approximations of $\pi$ and their applications
  • Investigate other series comparisons, such as $\sum \dfrac{1}{n^p}$ for different values of p
USEFUL FOR

Mathematicians, educators, students studying calculus, and anyone interested in series convergence and mathematical comparisons.

Albert1
Messages
1,221
Reaction score
0
$\sum \dfrac{1}{n^2}, \,\, and \,\,\,\dfrac{61}{36}$ which one is bigger ?
 
Mathematics news on Phys.org
Albert said:
$\sum \dfrac{1}{n^2}, \,\, and \,\,\,\dfrac{61}{36}$ which one is bigger ?

assumption n is from 0 to infinite
the 1st term = $\dfrac{\pi^2}{6} \lt \dfrac{10}{6}$
$\dfrac{10}{6} = \dfrac{60}{36} \lt \dfrac{61}{36}$
so second term is bigger
 
kaliprasad said:
assumption n is from 0 to infinite
the 1st term = $\dfrac{\pi^2}{6} \lt \dfrac{10}{6}$
$\dfrac{10}{6} = \dfrac{60}{36} \lt \dfrac{61}{36}$
so second term is bigger
very good solution Kaliprasad !
suppose $\sum \dfrac {1}{n^2} =\dfrac{\pi^2}{6} $ is not known yet , how can we compare those two numbers ?
 
Albert said:
very good solution Kaliprasad !
suppose $\sum \dfrac {1}{n^2} =\dfrac{\pi^2}{6} $ is not known yet , how can we compare those two numbers ?
solution:
$\sum \dfrac {1}{n^2}<1+ \dfrac{1}{2^2}+\dfrac {1}{3^2}+\dfrac {1}{3\times 4}+\dfrac {1}{4\times 5}+---+\dfrac {1}{(n-1)\times n}$
$=1+\dfrac {1}{4}+\dfrac{1}{9}+(\dfrac {1}{3}-\dfrac {1}{4})+(\dfrac {1}{4}-\dfrac {1}{5})+------+(\dfrac {1}{n-1}-\dfrac {1}{n})=\dfrac {61}{36}-\dfrac {1}{n}<\dfrac {61}{36}$
 

Similar threads

MHB Finding the Measure of Angle KPM Using Angle Bisector Theorem
  • · Replies 2 ·
Replies
2
Views
1K
MHB What is the measure of angle KPM?
  • · Replies 5 ·
Replies
5
Views
1K
MHB Comparing Trigonometric and Fractional Values: Which is Greater?
  • · Replies 4 ·
Replies
4
Views
2K
MHB Finding the Ratio of x,y,z in $R^+$
  • · Replies 2 ·
Replies
2
Views
1K
MHB Find Half a Number's Reciprocal Increased by Half its Reciprocal
  • · Replies 2 ·
Replies
2
Views
1K
MHB Roots of Polynomial: Find $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}$
  • · Replies 1 ·
Replies
1
Views
1K
MHB Can the Inequality of the Sum be Proven Using the Cube Root of -1?
  • · Replies 4 ·
Replies
4
Views
2K
MHB Can you prove this inequality challenge?
  • · Replies 1 ·
Replies
1
Views
1K
MHB What are the possible integers for which a given expression is an integer?
  • · Replies 1 ·
Replies
1
Views
1K
MHB Parentheses and brackets: Evaluate 2³[1/4+4(36÷12)]
  • · Replies 6 ·
Replies
6
Views
2K