I tried to solve this question, but can't get to a correct answer...

Let $a_n,b_n$ be two sequences s.t. $$ a_n\xrightarrow{n\to\infty}a>1,\quad b_n\xrightarrow{n\to\infty}b>1. $$ Find $$ \lim_{n\to\infty}\sqrt[n]{a_n^n+b_n^n+2018}$$

I tried to use sandwich, but it seems that the bounds I choose not working.

I checked in W|A for some examples and I think that the answer should be $\max\{a,b\}$.

Could someone hint me?

  • $\begingroup$ You recieved 4 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you. $\endgroup$ – 5xum Nov 14 '18 at 8:09
  • $\begingroup$ @5xum, will do, thanks for notifying :) $\endgroup$ – y12 Nov 14 '18 at 11:15


$$a_n^n + b_n^n + 2018 = a_n^n\left(1 + \left(\frac{b_n}{a_n}\right)^n + \frac{2018}{a_n^n}\right)$$

Now, use the fact that $\sqrt[n]{A\cdot B} =\sqrt[n]A \cdot \sqrt[n]B$ and you should be almost there. If $a>b$, then the inside of the parentheses goes to $1$, otherwise, it goes to $2$, but in either case, the $n$-th root of it converges to $1$.

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    $\begingroup$ $a > 1$ is given. Perhaps you want to assume that $a \ge b$? $\endgroup$ – Martin R Nov 12 '18 at 12:59
  • $\begingroup$ @MartinR Thanks. Yeah, that was my assumption. $\endgroup$ – 5xum Nov 12 '18 at 13:00
  • $\begingroup$ You can't assume $a>b$; they might be equal. $\endgroup$ – TonyK Nov 12 '18 at 13:32
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    $\begingroup$ @5xum: other than generalizing the result (and perhaps offering a better understanding), but as the question is stated, my comment is not relevant. $\endgroup$ – robjohn Nov 12 '18 at 14:02
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    $\begingroup$ @y12 Clearly, what you wrote is not true since on the RHS, the value of $n$ is not even defined. However, what you can use in finding the limit is that, so long as $x\in (1,\infty)$, you have $1<\sqrt[n]{x} <x$ for all $n\in\mathbb N$. $\endgroup$ – 5xum Nov 12 '18 at 14:31

A small issue:

Let $a \ge b$, and $n \ge n_0$ , s.t.

$a_n, b_n \gt 1.$


$ a_n(1+(b_n/a_n)^n +2018/a_n^n)^{1/n}.$

Note : For $n \ge n_0$

$(1+(b_n/a_n)^n +2018/a_n^n)$ is bounded.

Let $M_{n_0} >0$, real, be an upper bound.

$a_n < f(n) < a_n (M_{n_0})^{1/n}$.

Take the limit.

  • $\begingroup$ Such an $n_0$ does not necessarily exist if $a=b$. We might have $b_n>a_n$ for arbitrarily large $n$ (or for all $n$). Annoying, I know. $\endgroup$ – TonyK Nov 12 '18 at 13:29
  • $\begingroup$ TonyK.Quite annoying:)I'll try to fix it.Thanks $\endgroup$ – Peter Szilas Nov 12 '18 at 13:34
  • $\begingroup$ TonyK.Hopefully better now:)Thanks. $\endgroup$ – Peter Szilas Nov 12 '18 at 15:27
  • $\begingroup$ An alternative fix would be to note that $a_n^n+b_n^n+2018=A_n^n+B_n^n+2018$ where $A_n=\max(a_n,b_n)\to A=\max(a,b)$ and $B_n=\min(a_n,b_n)\to B=\min(a,b)$. $\endgroup$ – Barry Cipra Nov 12 '18 at 16:07
  • $\begingroup$ Barry Cipra.Very nice(!), thank you.Greetings. $\endgroup$ – Peter Szilas Nov 12 '18 at 16:26

We have that


and for $a\ge b>1$ wlog we obtain

$$\frac{\log\left(a_n^n+b_n^n+2018\right)}{n}=\log a_n+\frac{\log\left(1+\frac{b_n^n}{a_n^n}+\frac{2018}{a_n^n}\right)}{n} \to \log a$$

therefore the given limit is equal to $\max\{a,b\}$.

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    $\begingroup$ You can't assume $a>b$; they might be equal. Every answer so far has made this mistake! $\endgroup$ – TonyK Nov 12 '18 at 17:57
  • $\begingroup$ @TonyK Yes you are right, we need also to consider the case a=b, I fix that. $\endgroup$ – user Nov 12 '18 at 17:58
  • $\begingroup$ @TonyK Now it should be fixed including also the case with equality. The result doesn't change. Thanks to have pointed that out! Bye $\endgroup$ – user Nov 12 '18 at 18:04

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