How to evaluate $\lim_{n \to \infty} \frac1n \sum_{r=1}^n r^{\frac1r}$?
I've tried finding it, and I know that without the $\frac1n$ factor, the sequence has the limit $n$. What about the series? Will it be 1, then? How to show?
Asked
Active
Viewed 87 times
3
-
It cannot have the limit $n$ since $n\to\infty$. – JPi May 10 '14 at 13:23
-
1Apply the theorem of Cesaro-Stolz to reduce this to the limit of $n^{1/n}$, which is 1. – Lutz Lehmann May 10 '14 at 13:23
-
2@LutzL you should post this as hint-like answer – Norbert May 10 '14 at 13:27
-
@JPi, I didn't mean that. I've edited the question correctly. – A.Chakraborty May 10 '14 at 13:30
-
@LutzL, it actually reduces to the limit of $\frac{n^{\frac1n}}{n}$. Isn't it so? Now if we use product of limits, then it is 0. – A.Chakraborty May 10 '14 at 13:40
-
No, one has to represent the denominator also as a partial sum of n terms, and since $n-(n-1)=1$ ... – Lutz Lehmann May 10 '14 at 13:41
1 Answers
2
The theorem of Cesaro-Stolz states that if $b_n>0$ and $\sum_{n=1}^\infty b_n=\infty$, then $$ \lim_{n\to\infty}\frac{a_1+...+a_n}{b_1+...+b_n}=\lim_{n\to\infty}\frac{a_n}{b_n} $$ if the limit to the right exists. The left side may converge even if the right side does not.
This can be applied here fruitfully.
Lutz Lehmann
- 126,666