On limits related to $\frac1{\ln n}\sum\limits_{r=1}^{n^4}\frac1r$ when $n\to\infty$

Question

$$L=\lim_{n\to\infty}\frac1{\ln n}\sum_{r=1}^{n^4}\frac1r,\qquad M=\lim_{n\to\infty}\left\lfloor\frac1{\ln n}\sum_{r=1}^{n^4}\frac1r\right\rfloor$$

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I know that $$\lim_{n\to\infty}\frac1n\sum_{r=1}^{n^4}\frac nr=\int_{\lim\limits_{n\to\infty}1/n}^{\lim\limits_{n\to\infty}n^4/n}\frac{dx}x=\ln 4-\color{red}{\ln 0}\to-\infty$$

I suspect the limits to both doesn't exist.

Answer 1

This exercise is meant to help you discover/apply the properties of the partial sums of the harmonic series, known as the harmonic numbers $$H_n=\sum_{k=1}^n\frac1k.$$ For example, for every $n\geqslant2$, $$\log n\lt H_n\lt1+\log n.$$ This suffices to answer your question. Do you see why?

Note that, if $\lfloor\ \rfloor$ in the formula for $M$ stands for the integer part, the bounds above yield $$M=4.$$

Answer 2

Use the result

$$ \lim_{n \rightarrow \infty}\frac{1}{1}+\frac{1}{2}.....+\frac{1}{n} - \ln(n) = \gamma $$

This gives, $$\lim_{n \rightarrow \infty} \frac{4\ln(n)+\gamma}{\ln(n)} = 4$$

Answer 3

Regarding the first,

$$\lim_{n\to \infty} \sum_{r=1}^{n^4} \frac{1}{r} \sim \log (n^4) = 4 \log n$$

then

$$\lim_{n\to \infty} \left(\sum_{r=1}^{n^4} \frac{1}{r}/\log n\right)= 4$$

Answer 4

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$\tt\mbox{This is straightforward with Stoltz-Cesàro Theorem}$:

\begin{align}&\color{#66f}{\large% \lim_{n\ \to\ \infty}{\ds{\sum_{r\ =\ 1}^{n^{4}}}1/r \over \ln\pars{n}}} =\lim_{n\ \to\ \infty}{H_{n^{4}} \over \ln\pars{n}} =\lim_{n\ \to\ \infty}{\Psi\pars{n^{4} + 1} + \gamma \over \ln\pars{n}} \\[5mm]&=\lim_{n\ \to\ \infty} {\Psi\pars{\bracks{n + 1}^{4} + 1} - \Psi\pars{n^{4} + 1}\over \ln\pars{n + 1} - \ln\pars{n}} =\lim_{n\ \to\ \infty}n\braces{% \Psi\pars{\bracks{n + 1}^{4} + 1} - \Psi\pars{n^{4} + 1}} \\[5mm]&=\lim_{n\ \to\ \infty}n\braces{% \ln\pars{\bracks{n + 1}^{4} + 1} - \ln\pars{n^{4} + 1}} =\lim_{n\ \to\ \infty}n\ln\pars{\bracks{n + 1}^{4} + 1 \over n^{4} + 1} \\[5mm]&=\lim_{n\ \to\ \infty}n \ln\pars{1 + {4n^{3} + 6n^{2} + 4n \over n^{4} + 1}} =\lim_{n\ \to\ \infty}n \ln\pars{1 + {4 \over n}}=\color{#66f}{\Large 4} \end{align}