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.

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    $\begingroup$ 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? $\endgroup$ Commented Jun 27, 2021 at 16:48
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    $\begingroup$ What exactly is the $N$-th term test that you're talking about? $\endgroup$ Commented Jun 27, 2021 at 16:53
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    $\begingroup$ @AryamanMaithani I suspect it's what I know as the divergence test: if $a_n \not\to 0$, then $\sum a_n$ diverges. $\endgroup$ Commented Jun 27, 2021 at 16:54
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    $\begingroup$ @TheoBendit yes you are right. en.wikipedia.org/wiki/Term_test $\endgroup$
    – Broly-29
    Commented Jun 27, 2021 at 16:55
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    $\begingroup$ Notice when $n=1$, that it’s $c+d$ Which is greater than the limit of $d$, so it can’t be monotonically increasing. $\endgroup$
    – Eric
    Commented Jun 27, 2021 at 17:05

1 Answer 1


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 $-\infty$.

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    $\begingroup$ 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. $\endgroup$ Commented Jun 27, 2021 at 17:11
  • $\begingroup$ 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. $\endgroup$
    – Broly-29
    Commented Jun 27, 2021 at 17:15
  • $\begingroup$ 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 $\endgroup$ Commented Jun 27, 2021 at 17:19

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