Trigonometric Sum Challenge Σtan^(-1)(1/(n^2+n+1)=π/2

In summary, the conversation discusses the proof that tan^{-1}(k) can be expressed as a summation of tan^{-1}(\frac{1}{n^2+n+1}) and then deduces that the infinite sum of tan^{-1}(\frac{1}{n^2+n+1}) is equal to \frac{\pi}{2}. The proof utilizes induction and the assumption that tan^{-1}(j) can be expressed as the summation for k= j. The solution is appreciated by kaliprasad.
  • #1
lfdahl
Gold Member
MHB
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Show that

\[\tan^{-1}(k) = \sum_{n=0}^{k-1}\tan^{-1} \left ( \frac{1}{n^2+n+1} \right ),\;\;\;\;\; k \geq 1,\]

- and deduce that

\[ \sum_{n=0}^{\infty}\tan^{-1} \left ( \frac{1}{n^2+n+1} \right ) = \frac{\pi}{2}.\]
 
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  • #2
Looks like an obvious candidate for induction on ki. When k= 1, both sided are [tex]tan^{-1}(1)[/tex].

Assume that, for k= j, [tex]tan^{-1}(j)= \sum_{n=0}^{k-1} tan^{-1}\left(\frac{1}{n^2+ n+ 1}\right)[/tex]. Use that to show that
[tex]tan^{-1}(j+1)= \sum_{n=0}^{j} tan^{-1}\left(\frac{1}{n^2+ n+ 1}\right)[/tex].
 
  • #3
My solution

We have $n^2+n+1= 1+n(n+1) = \frac{1+n(n+1)}{(n+1) - n}$
Or $\frac{1}{n^2+n+1} = \frac{(n+1)-n}{1+(n+1)n}$
Using $\tan^{-1}\frac{a-b}{1+ab} = \tan^{-1} a - \tan^{-1}{b}$
We get $\tan^{-1} \frac{1}{n^2+n+1} = \tan ^{-1}(n+1)- \tan ^{-1}n$
Adding from 0 to k-1 we get as telescopic sum
Hence $\sum_{n=0}^{k-1} \tan^{-1} \frac{1}{n^2+n+1} = \tan ^{-1}k- \tan ^{-1}0 = \tan ^{-1}k$
Taking limit as $k = \infty$
$\sum_{n=0}^{\infty} \tan^{-1} \frac{1}{n^2+n+1} = \tan ^{-1}\infty= \frac{\pi}{2}$
 
  • #4
kaliprasad said:
My solution

We have $n^2+n+1= 1+n(n+1) = \frac{1+n(n+1)}{(n+1) - n}$
Or $\frac{1}{n^2+n+1} = \frac{(n+1)-n}{1+(n+1)n}$
Using $\tan^{-1}\frac{a-b}{1+ab} = \tan^{-1} a - \tan^{-1}{b}$
We get $\tan^{-1} \frac{1}{n^2+n+1} = \tan ^{-1}(n+1)- \tan ^{-1}n$
Adding from 0 to k-1 we get as telescopic sum
Hence $\sum_{n=0}^{k-1} \tan^{-1} \frac{1}{n^2+n+1} = \tan ^{-1}k- \tan ^{-1}0 = \tan ^{-1}k$
Taking limit as $k = \infty$
$\sum_{n=0}^{\infty} \tan^{-1} \frac{1}{n^2+n+1} = \tan ^{-1}\infty= \frac{\pi}{2}$

You´re truly a master, kaliprasad, thankyou very much for your nice solution!(Yes)
 

FAQ: Trigonometric Sum Challenge Σtan^(-1)(1/(n^2+n+1)=π/2

1. What is the "Trigonometric Sum Challenge Σtan^(-1)(1/(n^2+n+1)=π/2"?

The "Trigonometric Sum Challenge Σtan^(-1)(1/(n^2+n+1)=π/2" is a mathematical problem that involves finding the sum of an infinite series of trigonometric functions.

2. How do you solve the "Trigonometric Sum Challenge Σtan^(-1)(1/(n^2+n+1)=π/2"?

To solve the "Trigonometric Sum Challenge Σtan^(-1)(1/(n^2+n+1)=π/2", you need to use mathematical techniques such as telescoping series and trigonometric identities to simplify the series and find the value of the sum.

3. What is the significance of the sum being equal to π/2?

The significance of the sum being equal to π/2 is that it represents a special relationship between the values of the trigonometric functions tangent and arctangent. It also has applications in calculus and complex analysis.

4. Can the "Trigonometric Sum Challenge Σtan^(-1)(1/(n^2+n+1)=π/2" be generalized to other values?

Yes, the "Trigonometric Sum Challenge Σtan^(-1)(1/(n^2+n+1)=π/2" can be generalized to other values of the sum by changing the value of the expression inside the tangent function. This can lead to different sums depending on the value chosen.

5. What are the practical applications of the "Trigonometric Sum Challenge Σtan^(-1)(1/(n^2+n+1)=π/2"?

The "Trigonometric Sum Challenge Σtan^(-1)(1/(n^2+n+1)=π/2" has applications in mathematics, physics, and engineering. It can be used to solve various problems involving infinite series and trigonometric functions, as well as in the development of mathematical models and equations.

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