Determine if the sequence $a_n= (c^n +d^n)^\frac{1}{n}$ is an increasing or a decreasing sequence

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.

• How are you saying that the sequence is divergent? You've already said that it has $d$ as the limit. The test you're talking about seems to be one about the series $\sum a_n$ but no one is talking about that? Commented Jun 27, 2021 at 16:48
• What exactly is the $N$-th term test that you're talking about? Commented Jun 27, 2021 at 16:53
• @AryamanMaithani I suspect it's what I know as the divergence test: if $a_n \not\to 0$, then $\sum a_n$ diverges. Commented Jun 27, 2021 at 16:54
• @TheoBendit yes you are right. en.wikipedia.org/wiki/Term_test Commented Jun 27, 2021 at 16:55
• Notice when $n=1$, that it’s $c+d$ Which is greater than the limit of $d$, so it can’t be monotonically increasing.
– Eric
Commented Jun 27, 2021 at 17:05

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 $$-\infty$$.

• It should be pointed out that it only "must" be decreasing because we eliminated the other options. Simply starting from $c + d > d$ and converging to $d$ is not enough to conclude a sequence is monotone decreasing. Commented Jun 27, 2021 at 17:11
• yes i did it the same way. yes i got it that this is the decreasing sequence. But what about the nth term test ? we can't use it because it a sequence and not series. Commented Jun 27, 2021 at 17:15
• If nth term test is what you mean by divergence test, then I would say it is of no use, since you can already establish its converging Commented Jun 27, 2021 at 17:19