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$$\lim_{x\to 0}\left( \frac{1^x+2^x+3^x+\cdots+ n^x} n \right)^{a/x} $$

Michael Hardy
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Abhay Bagri
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  • I think the power should be $\frac {a}{x} $ – Archis Welankar Jun 23 '17 at 16:59
  • a/x is the power – Abhay Bagri Jun 23 '17 at 17:03
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    Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – Clement C. Jun 23 '17 at 17:09

3 Answers3

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Hint: Apply $\ln$ to get

$$a\cdot\frac{\ln (1^x + \cdots +n^x) - \ln n}{x}.$$

Let $f(x) = \ln (1^x + \cdots +n^x).$ Then the above equals

$$a\cdot\frac{f(x) - f(0)}{x-0},$$

which hopefully looks familiar.

zhw.
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  • Smart! Did you come up with this just now or were you familiar with similar problems that use this trick? – orlp Jun 23 '17 at 17:21
  • @orlp Whenever I see $0/0$ I first think "it's a derivative in disguise." Quite often it is. You just need to undisguise it. – zhw. Jun 23 '17 at 17:39
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we have

$$j^x=e^{x\ln (j)}=1+x\ln (j)+x\epsilon (x) .$$ then

$$S=\sum_{j=1}^nj^x=n+x\ln (n!)+x\epsilon (x).$$ and

$$\frac S n=1+\frac x n\ln (n!)+x\epsilon (x). $$ thus

$$(\frac S n)^\frac a x =e^{\frac a x \ln(1+\frac x n\ln (n!)+x\epsilon (x))}. $$

using the fact that $$\ln (1+X)\sim X \;\;(X\to 0) $$, we find the limit

$$\boxed {e^{\frac {a}{n}\ln (n!)}= (n!)^\frac a n }$$

hamam_Abdallah
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Case 1: $a\geq 0$

First; consider any decreasing positive sequence, say $(x_k)_{k=1}^{\infty}$ which decreases to $0$.

Using Jensen's inequality on concave function, we have

$$\frac{1+(2^{x_k})^{\frac{x_{k+1}}{x_k}}+(3^{x_k})^{\frac{x_{k+1}}{x_k}}+\dots +(n^{x_k})^{\frac{x_{k+1}}{x_k}}}{n}\leq \Big{(} \frac{1+2^{x_k}+3^{x_k}\dots +n^{x_k}}{n} \Big{)}^{\frac{x_{k+1}}{x_k}}$$

Hence $\Big{(} \frac{1+2^{x_k}+3^{x_k}\dots +n^{x_k}}{n} \Big{)}^{\frac{a}{x_k}}$ is a decreases as $x_k$ decreases.

Second, by AM-GM inequality,$$\Big{(} \frac{1+2^{x_k}+3^{x_k}\dots +n^{x_k}}{n} \Big{)}^{\frac{a}{x_k}}>(1\times2\times\dots\times n)^{\frac{x_k}{n}\frac{a}{x_k}}=(n!)^{\frac{a}{n}}$$.

Case 2: $a<0$

Everything will be same. Just the inequalities will reverse, then $\Big{(} \frac{1+2^{x_k}+3^{x_k}\dots +n^{x_k}}{n} \Big{)}^{\frac{a}{x_k}}$ is a increases as $x_k$ decreases.

These two cases will show that as $x\to 0^+$, the limit will be $(n!)^{\frac{a}{n}}$.

In the similar fashion you can prove that as $x\to 0^-$, the limit will be $(n!)^{\frac{a}{n}}$.

Hence, $$\lim_{x\to 0}\left( \frac{1^x+2^x+3^x+\cdots+ n^x} n \right)^{a/x}=(n!)^{\frac{a}{n}}\space \space \space \blacksquare $$

MAN-MADE
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  • Or, since $t\mapsto t^a$ is continuous, compute the limit without the $a$ and then raise the result to $a$. No need for a distinction of cases. – Clement C. Jun 23 '17 at 19:34
  • @ClementC. you are right, but I am not editing it 'cause it is marked by duplicate. – MAN-MADE Jun 24 '17 at 02:14