Adding arctan

Geometry Level 4

$\large \sum_{n=0}^\infty \text{arctan} \left( \dfrac1{n^2+n+1} \right)$

If the value of the summation above is in the form of $\dfrac da \pi ^y$ , where $a,d$ and $y$ are positive integers with $a,d$ coprime, find $d+a+y$ .

The answer is 4.

1 solution

Félix de Montemar
Mar 14, 2016

Since $\arctan { \left( \frac { 1 }{ { n }^{ 2 }+n+1 } \right) } =\arctan { \left( \frac { n+1-n }{ n\left( n+1 \right) +1 } \right) } =\arctan { \left( n+1 \right) } -\arctan { \left( n \right) }$ , this is a telescopic sum.

Then, $\sum _{ n=0 }^{ m }{ \left( \arctan { \left( n+1 \right) } -\arctan { \left( n+1 \right) } \right) } =\arctan { \left( m+1 \right) }$ and $\lim _{ m\rightarrow \infty }{ \sum _{ n=0 }^{ m }{ \arctan { \left( \frac { 1 }{ { n }^{ 2 }+n+1 } \right) } } } =\lim _{ m\rightarrow \infty }{ \arctan { \left( m+1 \right) } } =\frac { \pi }{ 2 }$ . So $d+a+y=1+2+1=4$

Did the exact same thing

Ραμών Αδάλια - 5 years, 3 months ago

Adding arctan

The answer is 4.

1 solution

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