# Given two convergent sequences find $\lim_{n\to\infty}\sqrt[n]{a_n^n+b_n^n+2018}$

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?

• 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. – 5xum Nov 14 '18 at 8:09
• @5xum, will do, thanks for notifying :) – y12 Nov 14 '18 at 11:15

Hint:

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

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 $$1$$.

• $a > 1$ is given. Perhaps you want to assume that $a \ge b$? – Martin R Nov 12 '18 at 12:59
• @MartinR Thanks. Yeah, that was my assumption. – 5xum Nov 12 '18 at 13:00
• You can't assume $a>b$; they might be equal. – TonyK Nov 12 '18 at 13:32
• @5xum: other than generalizing the result (and perhaps offering a better understanding), but as the question is stated, my comment is not relevant. – robjohn Nov 12 '18 at 14:02
• @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 $n\in\mathbb N$. – 5xum Nov 12 '18 at 14:31

A small issue:

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

$$a_n, b_n \gt 1.$$

$$f(n):=$$

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

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.

• 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. – TonyK Nov 12 '18 at 13:29
• TonyK.Quite annoying:)I'll try to fix it.Thanks – Peter Szilas Nov 12 '18 at 13:34
• TonyK.Hopefully better now:)Thanks. – Peter Szilas Nov 12 '18 at 15:27
• 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 $B_n=\min(a_n,b_n)\to B=\min(a,b)$. – Barry Cipra Nov 12 '18 at 16:07
• Barry Cipra.Very nice(!), thank you.Greetings. – Peter Szilas Nov 12 '18 at 16:26

We have that

$$\large{\sqrt[n]{a_n^n+b_n^n+2018}=e^{\frac{\log\left(a_n^n+b_n^n+2018\right)}{n}}}$$

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 $$\max\{a,b\}$$.

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