Evaluating a limit without usage of Riemann sum 
Evaluation of $$\lim_{n \rightarrow \infty}\bigg[\frac{1}{n}+\frac{1}{n+2}+\frac{1}{n+4}+\cdots \cdots +\frac{1}{3n}\bigg]$$

My work: Using Riemann sum
$$\lim_{n\rightarrow \infty}\sum^{n}_{r=0}\frac{1}{n+2r}=\lim_{n\rightarrow\infty}\sum^{n}_{r=0}\frac{1}{1+2\frac{r}{n}}\cdot \frac{1}{n}$$
Put $\displaystyle \frac{r}{n}=x$ and $\displaystyle \frac{1}{n}=dx$ and changing limits
$$\int^{1}_{0}\frac{1}{1+2x}dx=\frac{1}{2}\ln|1+2x|\bigg|^{1}_{0}=\frac{1}{2}\ln(3)$$
But would it be possible to do the problem without usage of the Rienmann sum? as in, could one solve such infinite sums in a method alternate to converting the sum into an integral.
 A: For even $n=2m$ we have
\begin{align}
\sum_{r=0}^{2m}\frac{1}{2m+2r} 
  &=\frac{1}{2}\sum_{r=0}^{2m}\frac{1}{m+r}=\\
  &=\frac{1}{2}\left(\sum_{r=1}^{3m}\frac{1}{r}-\sum_{r=1}^{m-1}\frac{1}{r}\right)=\frac{1}{2}(H_{3m}-H_{m-1}),
\end{align}
where
$$
H_n=\sum_{r=1}^n\frac{1}{r}
$$
are the Harmonic numbers.
Given the known relation
$$
\lim_{n\to\infty}(H_n-\log n)=\gamma
$$
we have
\begin{align}
&\lim_{m\to\infty}\sum_{r=0}^{2m}\frac{1}{2m+2r}
  =\frac{1}{2}\lim_{m\to\infty}(H_{3m}-H_{m-1})=\\
&\qquad=\frac{1}{2}\lim_{m\to\infty}[(H_{3m}-\log(3m))+\log(3m)-(H_{m-1}-\log(m-1))-\log(m-1)]=\\
&\qquad=\frac{1}{2}\lim_{m\to\infty}[\gamma+\log(3m)-\gamma-\log(m-1)]=\\
&\qquad=\frac{1}{2}\lim_{m\to\infty}\log\left(\frac{3m}{m-1}\right)=\frac{1}{2}\log 3.
\end{align}
For odd $n=2m+1$, taking into account
\begin{align}
&\frac{1}{n}+\frac{1}{n+2}+\ldots+\frac{1}{3n-2}+\frac{1}{3n}=\\
&=\left(\frac{1}{n}+\frac{1}{n+1}+\ldots+\frac{1}{3n-1}+\frac{1}{3n}\right)-\left(\frac{1}{n+1}+\frac{1}{n+3}+\ldots+\frac{1}{3n-3}+\frac{1}{3n-1}\right)
\end{align}
we can write
\begin{align}
\sum_{r=0}^{2m+1}\frac{1}{2m+1+2r}
  &= \sum_{s=0}^{4m+2}\frac{1}{2m+1+s}-\sum_{r=0}^{2m}\frac{1}{2m+2+2r}=\\
  &= \sum_{s=0}^{4m+2}\frac{1}{2m+1+s}-\frac{1}{2}\sum_{r=0}^{2m}\frac{1}{m+1+r}=\\
  &= H_{6m+3}-H_{2m}-\frac{1}{2}[H_{3m+1}-H_{m}]
\end{align}
and
\begin{align}
\lim_{m\to\infty}\sum_{r=0}^{2m+1}\frac{1}{2m+1+2r}
  &= \lim_{m\to\infty}\left(H_{6m+3}-H_{2m}-\frac{1}{2}[H_{3m+1}-H_{m}]\right)=\\
  &= \lim_{m\to\infty}\left(\log(6m+3)-\log(2m)-\frac{1}{2}[\log(3m+1)-\log(m)]\right)=\\
  &= \lim_{m\to\infty}\left(\log\left(\frac{6m+3}{2m}\right)-\frac{1}{2}\log\left(\frac{3m+1}{m}\right)\right)=\frac{1}{2}\log 3
\end{align}
Alternative proof
Let's rewrite the sum as
$$
\frac{1}{2}\sum_{r=0}^n\frac{1}{\frac{n}{2}+r}=\frac{1}{2}\left[\psi\left(\frac{3n+2}{2}\right)-\psi\left(\frac{n}{2}\right)\right],
$$
where $\psi$ is the digamma function and where we used the difference equation
$$
\psi(x+N)-\psi(x)=\sum_{k=0}^{N-1}\frac{1}{x+k},
$$
see Digamma::Recurrence formula and characterization.
Now, taking into account the following inequality, valid for $x>0$
$$
\log x-\frac{1}{x}\leq\psi(x)\leq\log x-\frac{1}{2x},
$$
see Digamma::Inequalities, we have
$$
\log\left(\frac{3n+2}{n}\right)-\frac{2}{3n+2}+\frac{1}{n}\leq\psi\left(\frac{3n+2}{2}\right)-\psi\left(\frac{n}{2}\right)\leq \log\left(\frac{3n+2}{n}\right)-\frac{1}{3n+2}+\frac{2}{n}
$$
and by the squeeze theorem, we get the result.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\Large\left. a\right)}$
\begin{align}
&\bbox[5px,#ffd]{\sum_{r = 0}^{n}{1 \over n + 2r}} =
\sum_{r = 0}^{n}\int_{0}^{1}t^{n + 2r - 1}\,\dd t =
\int_{0}^{1}\sum_{r = 0}^{n}t^{n + 2r - 1}\,\dd t
\\[5mm] = &\
\int_{0}^{1}t^{n - 1}\,{t^{2n + 2} - 1 \over t^{2} - 1}\,\dd t =
\int_{0}^{1}{t^{n - 1} - t^{3n + 1} \over 1 - t^{2}}\,\dd t =
{1 \over 2}\int_{0}^{1}{t^{n/2 - 1} - t^{3n/2} \over 1 - t}\,\dd t
\\[5mm] = &\
{1 \over 2}\pars{\int_{0}^{1}{1 - t^{3n/2} \over 1 - t}\,\dd t -
\int_{0}^{1}{1 - t^{n/2 - 1} \over 1 - t}\,\dd t} =
{H_{3n/2} - H_{n/2 -1} \over 2}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, &
{\bracks{\vphantom{\Large A}\ln\pars{3n/2} + \gamma + 1/\pars{3n}} -
\bracks{\vphantom{\Large A}\ln\pars{n/2 - 1} + \gamma +
1/\pars{n - 2}}\over 2}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\Large\to}\,\,\, &
\bbx{\ln\pars{3} \over 2} \\ &
\end{align}

$\ds{\Large\left. b\right)}$
\begin{align}
&\bbox[5px,#ffd]{\sum_{r = 0}^{n}{1 \over n + 2r}} =
\sum_{r = 0}^{\infty}\pars{{1 \over n + 2r} - {1 \over 3n + 2 + 2r}} \\[5mm] = &
{1 \over 2}\sum_{r = 0}^{\infty}\pars{{1 \over r + n/2} -
{1 \over r + 3n/2 + 1}} = \bbx{H_{3n/2} - H_{n/2 - 1} \over 2},
\quad\mbox{See}\ {\Large\left. a\right)}.\\ &
\end{align}
