Correct Answer!

By the :

f(x) =

arctan(x)

x²+1

for x ³ 1 is continuous, positive and decreasing since

æ
ç
è

arctan(x)

x²+1

ö
÷
ø

1-2(x)arctan(x)

( x²+1) ²

1-2(p/4)

( x²+1) ²

< 0

( )

Since

ó
õ

¥

1

arctan(x)

x²+1

dx =

ó
õ

p/2

p/4

u du =

æ
ç
è

ö
÷
ø

æ
ç
è

ö
÷
ø

converges, then by the , the series

¥
å
n = 1

| a_n| =

¥
å
n = 1

arctan(n)

n²+1

also converges. Hence, our original series, åa_n,.

Note that this problem can also be solved using the Direct Comparison Test and Limit Comparison Test.