Evaluating $\lim_{n\to \infty } \, \left(\sum _{k=1}^{\infty } \frac{1}{n}\right)$ $$\lim_{n\to \infty } \, \left(\sum _{k=1}^{\infty } \frac{1}{n}\right)$$
Intuitively, it seems that you are adding infinite of $\frac{1}{n}$ and then taking the limit as n goes to infinity, which would seem to give zero. 
Further, manipulating it as follows:
$$\sum _{k=1}^{\infty } \lim_{n\to \infty } \, \frac{1}{n}$$
gives that you are adding up an infinite number of zeroes, which would support the idea that the limit is zero.
Is this right?
 A: The answer is not defined, under the usual definitions of real analysis, or any definition of limits I know. (However, there may exist other definitions of limits that I'm unaware of, in settings larger than calculus or real analysis, and I'd guess that under definitions, the limit, if defined at all, would be $\infty$, whatever that means.)
In real analysis, we have definitions for $\lim_{n \to \infty} a_n$, where $a_n$ is a sequence of real numbers. Sometimes we use the symbol $\infty$ to denote that a sequence of real numbers grows arbitrarily large, and that therefore (in particular) the sequence has no limit in the real numbers.
However, what we have here is not the limit of a sequence of real numbers:
$$\lim_{n \to \infty} (\sum_{k=1}^{\infty} \frac1n) = \lim_{n \to \infty} a_n$$ where $a_n = \sum_{k=1}^{\infty} \frac1n$. Here, the expression $a_n$, itself an infinite sum (and therefore being a limit of finite sums, under the usual definitions of infinite sums) happens in this case to not be a real number:
$$a_n = \sum_{k=1}^{\infty} \frac1n = \lim_{m\to \infty} \sum_{k=1}^{m} \frac1n = \lim_{m\to\infty} \frac{m}n = \infty$$ (here I have used "$= \infty$" in the last step as notation to say that it is unbounded: does not exist in the real numbers).
So you are trying to find $\lim_{n\to\infty} a_n$, where $a_n$ is a limit that does not exist. If you want to allow "$\infty$" as an expression for manipulation, you could write $$\lim_{n \to \infty} \infty,$$ which is still not covered by the usual definitions.
But if you extend the definitions in any reasonable way to cover cases like this, I guess you'd define $\lim_{n \to \infty} \infty = \infty$. Note however that this is not standard.
A: $\lim_{n\to \infty } \, \left(\sum _{k=1}^{\infty } \frac{1}{n}\right)=\lim\limits_{n\to\infty}\infty=\infty$ since $\sum _{k=1}^{\infty } \frac{1}{n}=\infty\,.$
A: My hint:
I use Riemann sum integration:
$$\lim_{n\to \infty}\sum_{k=1}^n\frac{1}{k}=\lim_{n\to \infty}\frac{1}{n}\sum_{k=1}^n\frac{n}{k}=\int_0^1\frac{1}{x}dx=\lim_{x\to 0}\left(\ln 1-\ln x\right)=\infty$$
A: What is one to understand from $\lim_{n\to \infty } \, \left(\sum _{k=1}^{\infty } \frac{1}{n}\right)$?
is it $\lim_{n\to \infty } \, \left(\sum _{k=1}^{n } \frac{1}{n}\right)$ or $\lim_{n\to \infty } \, \left(\sum _{k=1}^{n^2 } \frac{1}{n}\right)$ or $\lim_{n\to \infty } \, \left(\sum _{k=1}^{\ln n } \frac{1}{n}\right)$?
Following part is added to show the difference between two of the above:
$\lim_{n\to \infty } \, \left(\sum _{k=1}^{n } \frac{1}{n}\right)$=1
$\lim_{n\to \infty } \, \left(\sum _{k=1}^{n^2 } \frac{1}{n}\right)=\infty$
