Limits of sums in which the sum index is a function of the variable involved in the limit I came across this problem when trying to prove that the harmonic series diverges.  I need to prove that:
\begin{equation}
\lim_{N \rightarrow \infty} \sum_{n=N+1}^{3N+1} \frac{1}{n} = 0
\end{equation}
Is he following methodology correct?
Firstly:
\begin{equation}
\lim_{N \rightarrow \infty} \sum_{n=N+1}^{3N+1} \frac{1}{n}
= \lim_{N \rightarrow \infty} \sum_{n=1}^{2N+1} \frac{1}{N + n} 
\end{equation}
Consider
\begin{equation}
\lim_{N \rightarrow \infty} \sum_{m=1}^{M} \frac{1}{N+m} \qquad \textrm{for some } M > 1, M \in \mathbb{N}
\end{equation}
but
\begin{equation}
\lim_{N \rightarrow \infty} \sum_{m=1}^{M} \frac{1}{N+m} 
= \lim_{N \rightarrow \infty} \frac{1}{N+1} + \lim_{N \rightarrow \infty} \frac{1}{N+2} + \space ... + \lim_{N \rightarrow \infty} \frac{1}{N+M} = 0 \space + 0 \space + \space ... + \space 0 = 0
\end{equation}
Hence is true that:
\begin{equation}
\lim_{N \rightarrow \infty} \sum_{m=1}^{M} \frac{1}{N+m} 
= 0 \qquad \forall \space M : M > 1, M \in \mathbb{N} 
\end{equation}
Set $M = 2N + 1$
\begin{equation}
\Rightarrow \lim_{N \rightarrow \infty} \sum_{m=1}^{2N+1} \frac{1}{N+m} 
= 0 \qquad N \in \mathbb{N} 
\end{equation}
\begin{equation}
\Rightarrow \lim_{N \rightarrow \infty} \sum_{n=N+1}^{3N+1} \frac{1}{n} = 0
\end{equation}
Update: 
The problem was misspecified.  It was required to prove 
\begin{equation}
\lim_{N \rightarrow \infty} \sum_{n=N+1}^{3N+1} \frac{1}{n}
\end{equation}
under the (false) assumption that the harmonic series has a finite limit $L$:
\begin{equation}
\lim_{N \rightarrow \infty} \sum_{n=1}^{N} \frac{1}{n} = L
\end{equation}
In this case (relying on the fact that the sum is monotonically increasing):
\begin{equation}
\lim_{N \rightarrow \infty} \sum_{n=1}^{N} \frac{1}{n} = L
\Rightarrow \nexists \space N : \sum_{n=1}^{N} \frac{1}{n} > L
\Rightarrow \nexists \space N : \sum_{n=1}^{3N+1} \frac{1}{n} > L
\Rightarrow \lim_{N \rightarrow \infty} \sum_{n=1}^{3N+1} \frac{1}{n} \le L
\end{equation}
but
\begin{equation}
\sum_{n=1}^{3N+1} \frac{1}{n} = \sum_{n=1}^{N} \frac{1}{n} +\sum_{n=N+1}^{3N+1} \frac{1}{n}
\end{equation}
\begin{equation}
\Rightarrow \lim_{N \rightarrow \infty} \sum_{n=1}^{3N+1} \frac{1}{n} = L + \lim_{N \rightarrow \infty} \sum_{n=N+1}^{3N+1} \frac{1}{n} \le L
\end{equation}
\begin{equation}
\Rightarrow \lim_{N \rightarrow \infty} \sum_{n=N+1}^{3N+1} \frac{1}{n} = 0
\end{equation}
Does that make sense?

The sum can be written as $H_{3N}-H_N$, where $H_n$ is the $n^{th}$ harmonic number. At the 
same time, we also have $H_N\simeq\ln N$, and $\ln a-\ln b=\ln\dfrac ab$. Combining the three, we 
conclude that the limit is $\ln3$. Your reasoning is incorrect, because, according to it, we 
could also write $~1~=~\lim\limits_{N\to\infty}\dfrac NN~=~\displaystyle\lim_{N\to\infty}\sum_{n=1}^N\dfrac1N~=~\lim_{N\to\infty}\dfrac1N~+~\lim_{N\to\infty}\dfrac1N~+~\cdots~+$ 
$+~\lim_{N\to\infty}\dfrac1N~=~0~+~0~+~\cdots~+~0~=~0,~$ which is absurd.
