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Calculate this limit $$\lim\limits_{n \to \infty} \frac{1}{n}\left(1+\frac{1}{\sqrt[n]{2}}+\frac{1}{\sqrt[n]{3}}+\dotsb+\frac{1}{\sqrt[n]{n}}\right).$$

I think inside the parentheses, each limit is $1$, and there are $n$ of them, so their sum is limited to $n$. Also,

$$\lim\limits_{n \to \infty}\frac{1}{n}=0.$$

Therefore I think, $$\lim\limits_{n \to \infty} \frac{1}{n}\left(1+\frac{1}{\sqrt[n]{2}}+\frac{1}{\sqrt[n]{3}}+\dotsb+\frac{1}{\sqrt[n]{n}}\right) = 0.$$

Is this solution correct? If so, how to prove it?

Did
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kdani
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    Hi, please always include some English words in the title, so that on right-clicking the title, the usual context menu will not be overridden by MathJax's menu. – user1551 May 03 '14 at 09:49

3 Answers3

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Actually the limit is $1$ because $\lim\limits_{n\to\infty}\sqrt[n]{n}=1$ and $$ \frac{n}{\sqrt[n]{n}}\leqslant1+\frac{1}{\sqrt[n]{2}}+\frac{1}{\sqrt[n]{3}}+\dotsb+\frac{1}{\sqrt[n]{n}}\leqslant n. $$ Note that your approach would also yield the limit $0$ for the sequence $$ \frac1n\cdot\left(1+1+\cdots+1\right), $$ for every number of terms in the parenthesis.

Did
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  • you write very fast i was just about to post the answer, nevertheless nice. – Shobhit May 03 '14 at 09:37
  • How about a more tricky question: instead of 1/2^(1/n) + 1/3^(1/n) + … we have 1/2^(1/2) + 1/3^(1/3) + … then all the values would not tend to 1 :)! Also, what can be said about having the harmonic series in the brackets (so 1/2 + 1/3 + 1/4), will that simply tend to 0? – Just_a_fool May 03 '14 at 09:42
  • @Just_a_fool Unsurprisingly, different settings might require different methods. But we knew that beforehand, didn't we? – Did May 03 '14 at 10:09
  • I'll add a link to the post about limit of $\sqrt[n]n$ (which is used here): http://math.stackexchange.com/questions/115822/how-to-show-that-lim-n-to-infty-n-frac1n-1 – Martin Sleziak Jan 27 '15 at 13:12
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Let $x_n = \dfrac{1}{n^{1/n}}$, then you can show : $x_n \to 1$ by various means, then apply Cesaro theorem to get the limit $1$.

DeepSea
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    If I am not mistaken, then by applying Stolz-Cesaro theorem we would get limit of slightly different expression: $\frac1n\left(1+\frac1{\sqrt[2]{2}}+\frac1{\sqrt[3]{3}}+\dotsb+\frac1{\sqrt[n]{n}}\right)$. – Martin Sleziak Jan 27 '15 at 13:09
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Using the Cezàro mean the desired limit is equal to $$\lim_{n\to\infty}\frac{1}{\sqrt[n]{n}}=1$$

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    If I am not mistaken, then by applying Stolz-Cesaro theorem we would get limit of slightly different expression: $\frac1n\left(1+\frac1{\sqrt[2]{2}}+\frac1{\sqrt[3]{3}}+\dotsb+\frac1{\sqrt[n]{n}}\right)$. – Martin Sleziak Jan 27 '15 at 13:09