Limit $\lim\limits_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+\cdots+\frac{n}{n^2+n}\right)$ $\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
Can we write it as following
$E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+2}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+3}\right)\cdots\cdots+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)\tag{1}$
Let's see what happens:-
$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)$$
$$\lim\limits_{n\to\infty}\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}=0$$
In the same way for further terms, we will get $0$
Let's also confirm for general term
$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)$$
$$\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)=0$$
So the whole expression $E$ will be zero
But actual answer is $1$
Let's see what happens if we evaluate the original expression $OE=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
$OE=\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{2}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{3}{n^2}}\cdots\cdots+\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)$
Now we can easily see that each term inside the bracket is tending to $0$, so can we say that sum of all terms upto infinity as well tends to zero?
I think we cannot because the quantity is not exactly zero, it is tending to zero, so when we add the values tending to zero upto infinity, we may not get zero.
But I got the following counter thought:-
$\lim\limits_{x\to0}\dfrac{(1+x)^\frac{1}{3}-1}{x}$
As we know $(1+x)^n=1+nx+\dfrac{n(n-1)}{2}x^2+\dfrac{n(n-1)(n-2)}{6}x^3\cdots\cdots\infty$ where $|x|<1$
$\lim\limits_{x\to0}\dfrac{\left(1+\dfrac{1}{3}x-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x^2+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^3\cdots\cdots\right)-1}{x}$
$\lim\limits_{x\to0}\dfrac{1}{3}-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^2\cdots\cdots$
Now here also all the terms except $\dfrac{1}{3}$ are tending to $0$. So here also we can say that the whole quantity may not turn out to be zero as we are adding all terms upto infinity. 
But surprisingly $\dfrac{1}{3}$ is the correct answer.
I am feeling very confused in these two things. Please help me.
 A: Observe that
$$\frac{n}{n^2+1}+\frac{n}{n^2+2}+\cdots+\frac{n}{n^2+n}$$ lies between
$$\frac{n}{n^2+n}+\frac{n}{n^2+n}+\cdots+\frac{n}{n^2+n}=\frac{n}{n+1}
$$ and
$$\frac{n}{n^2}+\frac{n}{n^2}+\cdots+\frac{n}{n^2}=1.$$
A: I will be giving a $\varepsilon -\mathcal{N}$ proof
notice
$$\begin{align}\frac{n}{1+n^2} + \frac{n}{2+n^2} + \cdots + \frac{n}{n+n^2} - 1=
n\Big( \frac{1}{1+n^2} - \frac{1}{n^2} +   \cdots +  \frac{1}{n+n^2} - \frac{1}{n^2}   \Big) = h_n\end{align}$$

Claim $\dfrac{n}{n+n^2} > \dfrac{r}{r+n^2}$


So
$|h_n| < \Big|n\cdot(\dfrac{1}{n^2}\cdot\dfrac{n^2}{n+n^2})\Big|=\dfrac{1}{n+1}<\dfrac{1}{n}$
so $\forall n > \dfrac{1}{\varepsilon}$
$|h_n| < \varepsilon$
A: That is not legitimate, because the number of terms inside the limit are growing as $n$ tends to infinity.
Rather, one sees that
\begin{align*}
\sum_{k=1}^{n}\dfrac{n}{n^{2}+n}\leq\sum_{k=1}^{n}\dfrac{n}{n^{2}+k}\leq\sum_{k=1}^{n}\dfrac{n}{n^{2}+1},
\end{align*}
the left and right-sided both tend to $1$.
A: If you know harmonic numbers
$$S_n=\sum_{i=i}^n \frac 1 {n^2+i}=H_{n^2+n}-H_{n^2}$$ Using the asympotics
$$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
   p^2}+O\left(\frac{1}{p^3}\right)$$ apply it twice and continue with Taylor series to get
$$n S_n=1-\frac{1}{2 n}-\frac{1}{6 n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit nd how it is approached.
Moreover, this gives a quite good approximation of the sum. Using $n=10$, the exact value is $\frac{11210403701434961}{11818204429243212} \approx 0.94857$ while the above truncated series gives $\frac{569}{600}\approx 0.94833$.
A: You can use the Squeeze Theorem, with
$$ n\cdot \frac{n}{n^2 + n} \leq a_n \leq n\cdot \frac{n}{n^2 + 1} $$
And as $ n \to \infty$, you get $  1 \leq \lim_{n \to \infty} a_n \leq 1 $, So, using the Squeeze Theorem, $a_n \to 1$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\sum_{k = 1}^{n}{n \over n^{2} + k}} =
n\sum_{k = 0}^{n - 1}{1 \over k + 1 + n^{2}}
\\[5mm] = &\
n\sum_{k = 0}^{\infty}\pars{{1 \over k + 1 + n^{2}} -
{1 \over k + n + 1 + n^{2}}} = n\pars{H_{n^{2} + n} - H_{n^{2}}}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,&
n\braces{\bracks{\ln\pars{n^{2} + n} + \gamma +
{1 \over 2n^{2} + 2n}} -
\bracks{\ln\pars{n^{2}} + \gamma + {1 \over 2n^{2}}}}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, &\
n\ln\pars{1 + {1 \over n}}
\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\,
\bbx{1} \\ &
\end{align}
$\ds{H_{z}}$ is a Harmonic Number and $\ds{\gamma}$ is the Euler-Mascheroni Constant.
A: Following the logic in your answer
$$
\lim_{n\to\infty}\overbrace{\left(\frac1n+\frac1n+\cdots+\frac1n\right)}^\text{$n$ terms}=0
$$
since each term tends to $0$. This is obviously false since the sum for each $n$ is $1$, so the limit is $1$.
We can say that a finite sum of terms which have a limit equals the finite sum of the limits. However, the same cannot be said for an unlimited sum of terms even though each has a limit. We must have some other theorem that allows us to swap order of the limits.
One proper way to handle the sum in the question is to bound the sum with
$$
\frac{n}{n+1}\le\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\cdots+\frac{n}{n^2+n}\right)\le1
$$
and then the Squeeze Theorem says the limit is $1$.
