If $0 \lt c\lt d$, then the sequence $a_n= (c^n +d^n)^\frac{1}{n}$
a.) is bounded and monotone decreasing.
b.) is bounded and monotone increasing.
c.) is monotone increasing , but unbounded for $1 \lt c\lt d$.
d.) is monotone decreasing , but unbounded for $1 \lt c\lt d$.
My approach:
We can clearly see $ \lim_{n\to \infty}$ $a_n =d $, hence it is bounded above so option c and d cannot be correct. Now using the Nth term test, $ \lim_{n\to \infty}$ $a_n =d $ ($d\ne 0$), we can say the sequence is divergent. But using this divergence can we say the sequence is monotonically increasing?
And if my approach is correct then c would be correct option.
Best Answer
I suppose you did like, $a_n= (c^n+d^n)^{\frac{1}{n}}= d\left(1+\left(\frac{c}{d}\right)^n\right)^\frac{1}{n}$, so $\lim a_n= d$. Now since limit exist its convergent. Its not divergent!
And also its monotone decreasing, since the first term is $c+d> d$ and as $n$ increases, $a_n\rightarrow d$, so it must be decreasing. (This approach is helpful only in mcq type exams)
Another way to show its decreasing: $a_{n+1}= d\left(1+\left(\frac{c}{d}\right)^{n+1}\right)^\frac{1}{n+1}< d\left(1+\left(\frac{c}{d}\right)^{n+1}\right)^\frac{1}{n}< d\left(1+\left(\frac{c}{d}\right)^{n}\right)^\frac{1}{n}$, since $c< d\Rightarrow \frac{c}{d}< 1\Rightarrow \left(\frac{c}{d}\right)\cdot\left(\frac{c}{d}\right)^n< \left(\frac{c}{d}\right)^n\Rightarrow \left(\frac{c}{d}\right)^{n+1}< \left(\frac{c}{d}\right)^n$, because $\frac{c}{d}> 0$.
Thus we have $a_{n+1}< a_n\,\,\,\forall n\in\mathbb{N}$
And also divergence not neccesarily imply the sequence is increasing, it may diverge to $-\infty$.