Question

Test the series whose general terms are
(1) $\sqrt{n^2+1}-n$
(2) $\sqrt{n^4+1}-\sqrt{n^4-1}$

Answer 1

$\left(1\right)$ $\sqrt{n^2+1}-n$

Let $a_n=\sqrt{n^2+1}-n$

Multiplying with $\sqrt{n^2+1}+n$ in numerator and denominator

$a_n=(\sqrt{n^2+1}-n)\times \frac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n}=\frac{n^2+1-n^2}{\sqrt{n^2+1}+n}=\frac{1}{\sqrt{n^2+1}+n}$

Let $b_n=\frac{1}{n^2}$

for all $n,0\le a_n\le b_n$

and $b_n$ converges when $n\to \infty$

${\sum }_{i=1}^{\infty }{b}_n={\sum }_{i=1}^{\infty }\frac{1}{n^2}=2{\sum }_{i=1}^{\infty }{a}_n<2$

Hence $a_n$ converges.

$\left(2\right)$ $\sqrt{n^4+1}-\sqrt{n^4-1}$

Let $a_n=\sqrt{n^4+1}-\sqrt{n^4-1}$

Multiplying and dividing $a_n$ by $\sqrt{n^4+1}+\sqrt{n^4-1}$

$a_n=\sqrt{n^4+1}-\sqrt{n^4-1}\times \frac{\sqrt{n^4+1}+\sqrt{n^4-1}}{\sqrt{n^4+1}+\sqrt{n^4-1}}=\frac{2}{\sqrt{n^4+1}+\sqrt{n^4-1}}$

Let $b_n=\frac{2}{n^2}$
for all $n,0\le a_n\le b_n$

and ${\sum }_{i=1}^{\infty }{b}_n=2\times 2=4$ (Converges)

${\sum }_{i=1}^{\infty }{a}_n<{\sum }_{i=1}^{\infty }{b}_n$

Hence $a_n$ converges.