Question

Which of the following is true for the sequence $\left\{\frac{1}{\sqrt[5]{5n^2+1}}\right\}$?

• A. The sequence is increasing and divergent.
• B. The sequence is increasing and convergent.
• C. The sequence is decreasing and divergent.
• D. The sequence is decreasing and convergent.
• E. The sequence is alternating and divergent.

Answer 1

The correct option is D The sequence is decreasing and convergent.

$limi{t}_{n\to \infty }\frac{1}{(n^2+1{)}^{1/5}}$
$=0$.
Hence as the value of $n$ increases, the value of function decreases. Thus it is a decreasing function.
Now applying radio test gives us
$limi{t}_{n\to \infty }|\frac{a_{n+1}}{a_n}|$
$=limi{t}_{n\to \infty }(|\frac{(n{)}^2+1}{(n+1{)}^2+1}|{)}^{1/5}$
$=limi{t}_{n\to \infty }(|\frac{n^2+1}{(n+1{)}^2+1}|{)}^{1/5}$
Now Denominator is greater than Numerator
$=limi{t}_{n\to \infty }(|\frac{n^2+1}{(n+1{)}^2+1}|{)}^{1/5}=0$
Hence if $r=limi{t}_{n\to \infty }(|\frac{n^2+1}{(n+1{)}^2+1}|{)}^{1/5}$ then $r<1$, Therefore, it is convergent.