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While playing around with some sequences, I came across the following problem:

Decide wether the limit of the sequence $\left(x_n\right)_{n=1}^{\infty}$ given by: $$ x_n=\frac{1}{n}\sum_{k=1}^{n}n^{\frac1k} $$ exits and if so, calculate it.

Under the assumption that the limit indeed exists, I proved that it is $2$:

Lower bound: $$ x_n=\frac{1}{n}\sum_{k=1}^{n}n^{\frac1k}=1+\frac{1}{n}\sum_{k=2}^{n}n^{\frac1k}≥1+\frac{n-1}{n}\implies\\ x:=\lim_{n\to\infty}x_n≥2 $$ For the upper bound I proved that: $$ x_{mn}≤\frac{1}{mn}\left[\left(\sum_{r=1}^{m-1}\exp\left(\frac{\log(mn)}{r}\right)\right)-(m-1)\exp\left(\frac{\log(mn)}{mn}\right)\right]+\left(mx_n\right)^\frac{1}{m} $$ Under the assumption that the limit exists, we can consider $n\to\infty$ which yields: $$ x≤1+\left(mx\right)^\frac{1}{m} $$ By then considering $m\to\infty$ this implies $x≤2$ and thus $x=2$.

But I cannot think of an approach to show the existence of $x$. The sequence seems to be monotone decreasing for $n≥13$ which suggests to use induction, but to relate $x_n$ and $x_{n+1}$ seems extremely hard. How can it be proven?

Thanks in advance.

Ps: If it helps, I can add the prove of the upper bound, just comment.

Redundant Aunt
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2 Answers2

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We can split the sum

\begin{align} \frac{1}{n}\sum_{k = 1}^n n^{\frac{1}{k}} &= 1 + \frac{1}{n}\sum_{2 \leqslant k \leqslant n^{1/3}} n^{\frac{1}{k}} + \frac{1}{n}\sum_{n^{1/3} < k \leqslant n} n^{\frac{1}{k}}\\ &\leqslant 1 + \frac{1}{n}\cdot \sqrt{n}\cdot n^{1/3} + \frac{n - \lfloor n^{1/3}\rfloor}{n}\cdot n^{1/(n^{1/3})}\\ &\leqslant 1 + n^{-1/6} + n^{1/(n^{1/3})}, \end{align}

where the second term converges to $0$, and the third to $1$ since $\frac{\log n}{n^{1/3}} \to 0$, showing

$$\limsup_{n\to\infty} \frac{1}{n}\sum_{k = 1}^n n^{\frac{1}{k}} \leqslant 2.$$

Daniel Fischer
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As an alternative to the fine solution already given, by integral test we have

$$1+n^\frac1n\le\frac{1}{n}\sum_{k=1}^{n}n^{\frac1k}\le 1+\int_1^n n^{\frac1x-1}dx =n^\frac1n-\frac{\log n}n\operatorname{Ei}\left(\frac{\log n}n\right)+\frac{\log n}n\operatorname{Ei}\left(\log n\right)$$

where $\operatorname{Ei}$ denotes the exponential integral function with

  • $1+n^\frac1n \to 2$
  • $\frac{\log n}n\operatorname{Ei}\left(\frac{\log n}n\right) \to 0 $
  • $\frac{\log n}n\operatorname{Ei}\left(\log n\right) \to 1$

therefore by squeeze theorem $\frac{1}{n}\sum_{k=1}^{n}n^{\frac1k} \to 2$.

To prove $\frac{\log n}n\operatorname{Ei}\left(\frac{\log n}n\right) \to 0 $ we ca use that as $x\to 0^+$

$$\operatorname{Ei}\left(x\right)= \gamma+\log x+O(x^2)\implies x\operatorname{Ei}\left(x\right)\to 0$$

To prove $\frac{\log n}n\operatorname{Ei}\left(\log n\right) \to 1$ we ca use that as $x\to \infty$

$$\operatorname{Ei}\left(x\right)= \frac{e^x}x+O(1/x)\implies \frac{\log n}n\operatorname{Ei}\left(\log n\right)=1+O(1/\log n) \to 1$$

user
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