Determine whether the sequence converges or diverges.If it converges, find the limit.

$a_{n}=\sqrt{\frac{1+4 n^{2}}{1+n^{2}}}$

Whether the sequence is converges or diverges and obtain the limit if the sequence is converges.

The sequence is converges and its limit is 2.

Given:

The sequence is an=1+4n21+n2 .

Definition used:

If an is sequence and limn→∞an is exists then the sequence an is convergent otherwise it is divergent.

Calculation:

Obtain the limit of the sequence to investigate whether the sequence is converges or diverges.

Compute limn→∞an=limn→∞1+4n21+n2 .

Divide the numerator and the denominator by the highest power.

limn→∞an=limn→∞1+4n2n21+n2n2=limn→∞1n2+4n2n21n2+n2n2

That is,

limn→∞an=limn→∞1n2+41n2+1

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

limn→∞an=limn→∞1n2+limn→∞(4)limn→∞(1n2)+limn→∞(1)=0+40+1=4=2

Since limn→∞an is exist and by using the definition of the sequence, it can be concluded that the sequence is converges to the limit 2.

Therefore, the sequence is converges to the limit 2.