# Evaluate a limit at infinity

Find the following limit without Lospital rule nor series $$\lim_{n\to \infty}\left(\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}\right)^{\frac{1}{n}}$$ I tried to take log to both sides to be $$\ln L=\lim_{n\to \infty}\frac{\ln\left((ab)^n +(bc)^n+(ac)^n\right)}{n}-\lim_{x\to \infty}\frac{n\ln(abc)}{n}$$ $$=\lim_{n\to \infty}\frac{\ln\left((ab)^n +(bc)^n+(ac)^n\right)}{n}-\ln(abc)$$ But i could not solve the first limit? $$0<a<b<c$$

• Something is not right. Commented Jan 9, 2018 at 14:37
• $n\to \infty$ ?? Commented Jan 9, 2018 at 14:39
• Surely all of the terms are constant with respect to $x$ and the limit is $$\left(\frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n} \right)^{\frac{1}{n}},$$ n'est-ce pas? Commented Jan 9, 2018 at 14:40
• The limit is w.r.t x Iam sorry Commented Jan 9, 2018 at 14:45
• Tell us where $a,b,c$ live please
– zhw.
Commented Jan 9, 2018 at 19:39

Hint:

If $0<a<b<c$, than we have $$\lim_{n \to \infty}\sqrt[n]{\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}}= \lim_{n \to \infty}\sqrt[n]{\frac{1}{a^n}}=\frac{1}{a}$$

• If $0<a<b<c<1$ the result is not the inverse of the less but the inverse of the greatest. Can you see why? Commented Jan 9, 2018 at 16:53
• @samjoe: Sorry! You are right and I made a stupid mistake ! I edit..:) Commented Jan 9, 2018 at 19:37
• This clearly is simplest approach! Commented Jan 10, 2018 at 2:51

I don't know if it's mathematically right but if you take the limit: $$\lim_{n\to \infty}\left(\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}\right)^{\frac{1}{n}}$$ And put the terms on the same denominator

$$\lim_{n\to \infty}\left(\frac{(bc)^n+(ac)^n+(ab)^n}{(abc)^n}\right)^{\frac{1}{n}} = \frac{1}{abc} \lim_{n\to \infty}\left((bc)^n+(ac)^n+(ab)^n\right)^{\frac{1}{n}}$$ This limit doesn't diverge becuase what is inside the limit is (I think) smaller or equal than $$bc + ac+ ab$$ So a upper bound of the limit is $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$$ Correct me if I'm wrong, I'm not 100% sure.

• I think what inside is not smaller than or equall $$bc+ac+ab$$nut also iam not sure @WarreG Commented Jan 9, 2018 at 15:14
• It's true for $n=1$ and for $n=2$ it's pretty much $$\sqrt{a^2+b^2} \leqslant a+b$$ and that is also true. That was my reasoning, you might be able to prove it by induction but that would bring it too far. Commented Jan 9, 2018 at 15:27
• Nope, I'm wrong.You need to have more information about a b and c. Commented Jan 9, 2018 at 15:29
• I have mentioned informations about a , b and c Commented Jan 9, 2018 at 20:31

Assume $a,b,c>0$ and wlog $\frac1a=max\{\frac1a,\frac1b\frac1c\}$ thus

$$\left(\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}\right)^{\frac{1}{n}}=e^{\frac{\log{\left(\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}\right)}}{n}}=e^{\frac{\log{\left(\frac{1}{a^n}\right)}+\log{\left(1+\frac{a^n}{b^n}+\frac{a^n}{c^n}\right)}}{n}}\to\frac1a$$

indeed

$$\frac{\log{\left(\frac{1}{a^n}\right)}+\log{\left(1+\frac{a^n}{b^n}+\frac{a^n}{c^n}\right)}}{n}=\frac{n\log{\left(\frac{1}{a}\right)}+\log{\left(1+\frac{a^n}{b^n}+\frac{a^n}{c^n}\right)}}{n}=$$ $$=\log{\left(\frac{1}{a}\right)}+\frac{\log{\left(1+\frac{a^n}{b^n}+\frac{a^n}{c^n}\right)}}{n}\to \log{\left(\frac{1}{a}\right)}+0=\log{\left(\frac{1}{a}\right)}$$

• How the second limit in the last row tends to 0 ? @gimusi Commented Jan 9, 2018 at 16:24
• Since $0\leq\log{\left(1+\frac{a^n}{b^n}+\frac{a^n}{c^n}\right)}\leq\log 3$ then $$\frac{\log{\left(1+\frac{a^n}{b^n}+\frac{a^n}{c^n}\right)}}{n}\to 0$$
– user
Commented Jan 9, 2018 at 16:33

Assume $a,b,c>0.$ Let $M= \max (a,b,c).$ Then

$$M = (M^n)^{1/n} < (a^n+b^n + c^n)^{1/n} \le (3M^n)^{1/n} = 3^{1/n}M.$$

Since $3^{1/n} \to 1,$ we see $(a^n+b^n + c^n)^{1/n} \to M.$ The answer for the problem as stated is therefore $\max (1/a,1/b,1/c).$