# 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. – Michael Rozenberg Jan 9 '18 at 14:37
• $n\to \infty$ ?? – jonsno Jan 9 '18 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? – Xander Henderson Jan 9 '18 at 14:40
• The limit is w.r.t x Iam sorry – Hussien Mohamed Jan 9 '18 at 14:45
• Tell us where $a,b,c$ live please – zhw. Jan 9 '18 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? – Emilio Novati Jan 9 '18 at 16:53
• @samjoe: Sorry! You are right and I made a stupid mistake ! I edit..:) – Emilio Novati Jan 9 '18 at 19:37
• This clearly is simplest approach! – jonsno Jan 10 '18 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 – Hussien Mohamed Jan 9 '18 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. – WarreG Jan 9 '18 at 15:27
• Nope, I'm wrong.You need to have more information about a b and c. – WarreG Jan 9 '18 at 15:29
• I have mentioned informations about a , b and c – Hussien Mohamed Jan 9 '18 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 – Hussien Mohamed Jan 9 '18 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 Jan 9 '18 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 $\max (1/a,1/b,1/c).$