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Find the value of $$\lim_{n\rightarrow \infty}\frac{1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n^3}}{\ln(n)}$$

My Try: Using Stolz-Cesaro,

Let $\displaystyle a_{n} = 1+\frac{1}{2}+\frac{1}{3}+\cdots \cdots +\frac{1}{n}$ and $b_{n} = \ln(n)$

So $\displaystyle \frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} = \lim_{n\rightarrow \infty}\frac{1}{{(n+1)^3}}\cdot \frac{1}{\ln\bigg(1+\frac{1}{n}\bigg)} = 0$

Please explain if what I have done above is right.

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  • $\begingroup$ I haven't heard of Cizero stolza but I would bound the numerator by $\int_1^{n} \frac 1 {x^{3}} \, dx =\frac 1 {2n^{2}}$ to show that the limit is $0$. $\endgroup$ – Kavi Rama Murthy Sep 10 '18 at 8:51
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    $\begingroup$ I wonder why so many downvotes. There was a simple mistake in asker's application of Cesaro-Stolz, but he did show effort. +1 to compensate. $\endgroup$ – Paramanand Singh Sep 10 '18 at 13:26
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Let $H_n=\sum_{k=1}^n \frac 1k$. Since $H_n=\log n +O(1)$, $$H_{n^3} = 3\log n +O(1)$$ thus$$\frac{H_{n^3}}{\log n} = 3+O\left(\frac{1}{\log n} \right)=3+o(1)$$

The limit is $3$.

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Use of Cesaro-Stolz is tricky here. Let's first evaluate the limit of $a_n$ given by $$a_n=\frac{1}{\log n} \sum_{k=1}^{n}\frac{1}{k}$$ Using Cesaro-Stolz we have $$\lim_{n\to \infty} a_n=\lim_{n\to\infty} \frac{1/n}{\log n-\log(n-1)}=-\lim_{n\to \infty} \frac{1}{n\log(1-(1/n))}=1$$ and then the sequence in question, say $b_n$, is given by $b_n=3a_{n^3}$ so $b_n\to 3$ as $n\to\infty$.

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  • $\begingroup$ I like this creativity! (+1) $\endgroup$ – Shashi Sep 10 '18 at 11:08
  • $\begingroup$ Great simplification! $\endgroup$ – gimusi Sep 10 '18 at 12:15
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Your $a_n$ should go up to $\frac1{n^3}$. Then $a_{n+1}-a_n=\frac{1}{(n+1)^3}+\frac{1}{(n+1)^3-1}\dots+\frac1{n^3+2}+\frac{1}{n^3+1}$.

I don't know whether you can arrive there by your method, but you can find the limit by using the asymptotics of harmonic numbers / definition of the euler $\gamma$ constant. Then the limit should be $3$.

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Note that using Stolz-Cesaro we obtain

$$\frac{\sum_{k=1}^{(n+1)^3}\frac1k-\sum_{k=1}^{n^3}\frac1k}{\ln(n+1)-\ln n}\sim\frac{\ln(n+1)^3-\log n^3}{\ln(1+\frac1n)}=3\frac{\ln(1+\frac1n)}{\ln(1+\frac1n)}=3$$

but, as already noted by Gabriel Romon, this is not necessary and it is sufficient use that by Harmonic series

$$\frac{\sum_{k=1}^{n^3}\frac1k}{\ln n}\sim\frac{\ln n^3}{\ln n}=3$$

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    $\begingroup$ A direct use of Cesaro-Stolz leads to complications. I have tried a slight variation using subsequences. $\endgroup$ – Paramanand Singh Sep 10 '18 at 11:05
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To answer your question directly. The mistake is made in $a_{n+1}-a_n$. For example, if $n = 10$, then this is given by $$a_{11}-a_{10} = 1/1331 + 1/1330 + \cdots + 1/1001$$ since $11^3 = 1331$ and $10^3=1000$.

If you insert this, you still are going to need to do something similar as Gabriel suggested by viewing the summation as upper and lower Riemann sums.

Edit2: if you now use Stolz Cesaro, which is possible, it works out as Gimusi did.

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  • $\begingroup$ Why cannot we apply CS? $\endgroup$ – gimusi Sep 10 '18 at 10:04
  • $\begingroup$ I am sorry you're right. Made a mistake here. $\endgroup$ – Stan Tendijck Sep 10 '18 at 10:14

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