**Given:**

The sequence is an=arctan(n2n2+4).

**Definition used:**

If an is a sequence and limn→∞an exists, then the sequence an is said to be convergent otherwise it is divergent.

**Result used: Limit chain rule**

If limu→bf(u)=L and limx→ag(x)=b with f(x) is continuous at x=b, then the value of limx→af(g(x)) is *L*.

**Calculation:**

Obtain the first 20 terms of the sequence.

*n* | n2n2+4 | an=arctan(n2n2+4) |

1 | 15=0.2000 | arctan(0.2000)=0.2027 |

2 | 48=0.5000 | arctan(0.5000)=0.4636 |

3 | 913=0.6923 | arctan(0.6923)=0.6055 |

4 | 1620=0.8000 | arctan(0.8000)=0.6747 |

5 | 2529=0.8621 | arctan(0.8621)=0.7115 |

6 | 3640=0.9000 | arctan(0.9000)=0.7328 |

7 | 4953=0.9245 | arctan(0.9245)=0.7462 |

8 | 6468=0.9412 | arctan(0.9412)=0.7551 |

9 | 8185=0.9529 | arctan(0.9529)=0.7613 |

10 | 100104=0.9615 | arctan(0.9615)=0.7658 |

11 | 121125=0.9680 | arctan(0.9680)=0.7691 |

12 | 144148=0.9730 | arctan(0.9730)=0.7717 |

13 | 169173=0.9768 | arctan(0.9768)=0.7737 |

14 | 196200=0.9800 | arctan(0.9800)=0.7753 |

15 | 225229=0.9825 | arctan(0.9825)=0.7766 |

16 | 256260=0.9846 | arctan(0.9846)=0.7776 |

17 | 289293=0.9863 | arctan(0.9863)=0.7785 |

18 | 324328=0.9878 | arctan(0.9878)=0.7793 |

19 | 361365=0.9890 | arctan(0.9890)=0.7799 |

20 | 400404=0.9891 | arctan(0.9901)=0.7799 |

Plot the points (n,an), for n=1,2,...20 on the graph as shown below in Figure 1.

From the graph, it is observed that the sequence is convergent.

Also, the value of the limit can be guessed to lies between 0.7 and 0.8 as the plotted points are closer to between 0.7 and 0.8.

**To prove:**

The sequence an=arctan(n2n2+4) converges.

**Proof:**

Obtain the limit of the sequence (the value of the term an as *n* tends to infinity).

That is, compute limn→∞an=limn→∞(arctan(n2n2+4)).

Let g(n)=n2n2+4 and f(u)=arctan(u).

Obtain limn→∞g(n).

limn→∞g(n)=limn→∞(n2n2+4)

Divide the numerator and the denominator by the highest power.

limn→∞(n2n2+4)=limn→∞(n2n2n2+4n2)=limn→∞(1n2n2+4n2)limn→∞(11+4n2)

Redefine the terms as follows:

limn→∞(n2n2+4)=limn→∞(1)limn→∞(1)+limn→∞(4n2)=11+limn→∞(4n2)

Apply infinity property limn→∞(kna)=0 and simplify the terms.

limn→∞(n2n2+4)=11+1∞=11+0=11=1

Obtain limu→1f(u).

limu→1f(u)=limu→1(arctan(u))=arctan(1)=π4≈0.7854

Apply the limit chain rule and obtain the limit.

Since g(n)=n2n2+4 and f(u)=arctan(u), limn→∞(n2n2+4)=π4 and limn→∞(arctan(n2n2+4))=π4.

Therefore, limn→∞f(g(n))=π4.

Since limn→∞an exists and by using the definition, it can be concluded that the sequence is converging to the limit π4.

Therefore, the sequence is convergent to the limit π4≈0.7854.