The limit of $u_n=\sum_{i=1}^n\frac{1}{i}$. I had this question : 
Does the sequence 
$$\begin{align*}
u_n=\sum_{i=1}^n\frac{1}{i}
\end{align*}$$
have a real end ? 
Well my teacher said no but the last question was 
Is 
$$\begin{align*}
u_n=\sum_{i=1}^n\frac{1}{i^2}
\end{align*}$$
a convergent sequence ? 
And we proved yes ...
I don't know maybe there is a problem in my understanding 
Can a convergent sequence have no real limit ?
and if the first doesn't converge why doesn't it ? can it be proven in a similar way to the second?
 A: The sequence $(u_n)_n$ defined by $$u_n=\sum_{i=1}^n\frac1i$$
does not converge. To see why, you can look up harmonic series
The sequence $(v_n)_n$ defined by $$v_n=\sum_{i=1}^n\frac1{i^2}$$
does converge.
They are not the same sequence, so there's no problem there.

The answer to your last question:

Can a convergent sequence have no real limit ?

is NO. A sequence is convergent, by definition, if it has a limit.
A: Let's first get our definitions straight: 
The following is a partial or finite sum:
$$s_n=\sum_{k=1}^na_k.$$
We call the infinite series (or infinite sum) $\sum_{k=1}^∞a_k$ convergent with limit $s_∞$, if its sequence of partial sums converges:
$$s_n=\sum_{k=1}^na_k → s_∞\qquad (n→∞).$$
The Cauchy-criterion for sequences states, that  an infinite sum is convergent if for all $ε>0$ there exists a $n_ε∈ℕ$, such that for all $n>m≥n_ε$ it holds:
$$|s_n-s_m|=\left|\sum_{k=m+1}^na_k\right|<ε.$$

The first infinite series you stated is
$$\sum_{k=1}^∞\frac{1}{k},$$
the so called harmonic series. 
For this series it holds
$$s_{2m}-s_m = \sum_{k=m+1}^{2m} \frac{1}{k} = \frac{1}{m+1} + …+\frac{1}{2m} > \underbrace{\frac{1}{2m}+…+\frac{1}{2m}}_{m \text{ times}} = \frac{1}{2}.$$ 
Hence the Cauchy criterion is not fulfilled and the series is not convergent. 

The second infinite sum is $$\sum_{k=1}^∞\frac{1}{k^2}.$$
That sum is convergent, but we need the following lemma to help us proof that.
Lemma: Given two series $s_∞=\sum a_k$ and $s_∞'=\sum a_k'$, if for almost all $k∈ℕ$ it holds $$|a_k|≤κa_k',$$ 
with some constant $κ$, then the absolute convergence of $s_∞'$ ensures the convergence of $s_∞$.
Now we can use that. It is
$$\frac{1}{k^2} = \frac{1}{k(k+1)} + \frac{1}{k^2(k+1)} ≤ \frac{2}{k(k+1)}.$$
And it holds:
$$\sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{k=1}^n[\frac{1}{k}-\frac{1}{k+1}] = 1 - \frac{1}{n+1}$$
Hence  $$\sum_{k=1}^n\frac{1}{k(k+1)} → 1\qquad (n→∞),$$
und thus $\sum_{k=1}^∞\frac{1}{k^2}$ is convergent.


Can a convergent sequence have no real limit ?

The answer is no. One of the ways real numbers can be described is by construction from Cauchy sequences. (see here)
The motivation for that construction is, that convergent sequences with elements in $ℚ$ not necessarily have their limit in $ℚ$. 
By the construction $ℝ$ is the completion of $ℚ$. That means every Cauchy sequence (with elements in $ℝ$) converges in $ℝ$. 
A: Suppose that $u_n$ converges thus $u_n$ is a cauchy sequence, i.e:
$\forall \epsilon >0 ,\exists n_0 \in \mathbb{N}$ such that $|u_n-u_m|< \frac{1}{2}, \forall m,n \geqslant n_0$
Let $m>n \geqslant n_0$
$|u_m-u_n|=|\frac{1}{n+1}+...+ \frac{1}{m}|$
For $m=2n$ we have that  $$\frac{1}{2}= \frac{n}{2n} \leqslant \frac{1}{2n}+...+\frac{1}{2n} \leqslant \frac{1}{n+1}+...+ \frac{1}{2n}< \frac{1}{2}$$ 
which is a contradiction.
$$Second-proof$$
We will prove that exists $n_0 \in \mathbb{N}$ such that $\log{(1 +\frac{1}{n})} \leqslant \frac{2}{n}$
This is easy because $n\log{(1 +\frac{1}{n})} \rightarrow \log{e}=1<2$ 
thus $\exists n_0 \in \mathbb{N}$ such that $\forall n \geqslant n_0$ we have that $$\frac{1}{n} \geqslant \frac{1}{2} \log{(1 +\frac{1}{n})}= \frac{1}{2}\log{(\frac{n+1}{n})}= \frac{1}{2}(\log{(n+1)}- \log{n})$$
Finally we have that $\forall n \geqslant n_0$ $$\sum_{i=1}^n \geqslant \frac{1}{2} \sum_{i=1}^n (\log{(i+1)}-\log{i})= \frac{1}{2} \log{(n+1)}$$
Take limits and you have your conclusion.
For the second part of your question use the fact that $\forall n \geqslant 2$
$$ \frac{1}{n^2} \leqslant \frac{1}{(n-1)n} =\frac{1}{n-1}- \frac{1}{n}$$
I hope this helps.
