Does $\lim_{n\to\infty} \sum^{n^2}_{k=n}\frac{1}{k}$ exist? Does $\lim_{n\to\infty}\sum^{n^2}_{k=n}\frac{1}{k}$ exist?
MY TRIAL 
\begin{align} \lim_{n\to\infty}\sum^{n^2}_{k=n}\frac{1}{k}=\int^{n^2}_{n}\frac{1}{t} \,\mathrm{d}t=\ln n^2- \ln n=\ln n\end{align}
Please, am I right? If no, can anyone show me the right answer? Thanks! 
 A: Demo. 2 methods. 


*

*Using the asymptotic expression of harmonic series
$$
H_n = \sum_1^n \frac 1k = \log(n) + \gamma + e_n\quad [e_n \to 0 \text{ as } n \to \infty],
$$
where $\gamma$ is the Euler-Mascheroni constant, we have
$$
\sum_n^{n^2} \frac 1k = H_{n^2} - H_{n-1} = 2\log(n) - \log(n-1) + e_{n^2} - e_{n-1} \geqslant \log(n-1) + e_{n^2} - e_{n-1},
$$
let $n \to \infty$, then the limit is $+\infty$.

*Since $1/x$ is decreasing on $(0, +\infty)$, we have 
$$
\begin{split}
\sum_n^{n^2} \frac 1k 
 &= \sum_n^{n^2} \int_{k}^{k+1}\frac 1k \mathrm d x \\
 &\geqslant \sum_n^{n^2} \int_{k}^{k+1}\frac 1x \mathrm d x \\
 &= \int_n^{n^2+1} \frac 1x \mathrm d x \\
 &= \log(n^2 +1) - \log(n) \\
 &> \log(n^2) - \log (n) \\
 &= \log (n) \to +\infty
\end{split}
$$
hence the limit is $+\infty$. 

A: Actually you do not need integrals to prove that your limit is $+\infty$. By letting
$$ T_n = \sum_{k=n}^{n^2}\frac{1}{k} = \frac{1}{n^2}+\sum_{j=1}^{n-1}\sum_{k=jn}^{(j+1)n-1}\frac{1}{k}$$
we have
$$ T_n \leq \frac{1}{n^2}+\sum_{j=1}^{n-1}n\cdot\frac{1}{jn}=\frac{1}{n^2}+H_{n-1} $$
as well as
$$ T_n \geq \frac{1}{n^2}+\sum_{j=1}^{n-1}n\cdot\frac{1}{(j+1)n}=\frac{1}{n^2}-1+H_n$$
so you may draw the conclusion from the divergence of the harmonic series.
With integrals a correct approach is given by
$$ T_n = \int_{n}^{n^2+1}\frac{dx}{\lfloor x\rfloor}\geq\int_{n}^{n^2+1}\frac{dx}{x}\geq\int_{n}^{n^2}\frac{dx}{x}=\log n. $$
A: Your idea is absolutely correct, just some details are wrong.
For $t\ge k$ we have $\frac1 t \le \frac1k$. So $$\sum_{k=n}^{n^2-1}\frac1k \ge \sum_{k=n}^{n^2-1} \int_{k}^{k+1} \frac1tdt = \int_n^{n^2} \frac1tdt = \ln n.$$ Hence $$\sum_{k=n}^{n^2} \frac1k \ge \ln n +\frac1{n^2},$$
which tends to infinity as $n\to\infty$.
A: $\tiny{An\ option:}$
$\tiny{\sum_0=\dfrac{1}{n}}$.
$\tiny{\sum_1= \dfrac{1}{n+1}+....+\dfrac{1}{2n} \gt
n\dfrac{1}{2n}}$.
$\tiny {\sum_2= \dfrac{1}{2n+1} +\dfrac{1}{2n+2} +.. \dfrac{1}{3n} \gt n\dfrac{1}{3n}}$.
$\tiny {\sum_{n-1}= \dfrac{1}{(n-1)n+1}+.....\dfrac{1}{(n-1)n +2}+.....+}$
$\tiny{\dfrac{1}{n(n-1)+n} \gt n\dfrac{1}{n^2}}.$
$\tiny{Combining:}$
$\tiny {\sum_{k=n}^{n^2} \dfrac{1}{k}=\sum_{0}+\sum_{1}....+ \sum_{n-1} \gt}$
$\tiny{ 1/n +(1/2+1/3 +1/4+.....1/n )}$.
