Limit inferior and inequality Suppose you have a constant $d$ and a sequence $(a_n)_n$ such that we have $d < \liminf a_n$. Note that we do not assume that $a_n$ is even bounded. I want to prove that there exists a constant $C$ such that $\sum_{k = 1}^{n-1}\frac{a_k}{k} \geq C + \sum_{k = 1}^{n-1} \frac{d}{k}$ holds for all $n \in \mathbb{N}$. I think you have to use a crucial property of the limit inferior but I'm unable to figure it out by myself.
I'd appreciate every help.
 A: There exists an $N$ such that $a_n > d$ for all $n \ge N$. [Otherwise, you can obtain a subsequence $(a_{n_k})_k$ such that $a_{n_k} \le d$ for all $k$, which will then have $\liminf_k a_{n_k} \le d < \liminf_n a_n$, a contradiction.]
Then taking $C = \min_{n \in \{1,\ldots,N-1\}}\sum_{k=1}^n \frac{a_k - d}{k}$ (possibly negative) gives you what you want, since
$$\sum_{k=1}^n \frac{a_k - d}{k} \ge C$$
for $1 \le n \le N - 1$ automatically, and moreover
$$\sum_{k=1}^n \frac{a_k - d}{k} \ge \sum_{k=1}^{N-1} \frac{a_k - d}{k} \ge C$$
for $n \ge N$ since $a_k > d$ when $k \ge N$.
A: Define $\alpha:=\liminf a_n$. For each $\varepsilon>0$ exists $N\in\mathbb N$ such that $a_n\geq\alpha-\varepsilon$ for $n\geq N$. Assume not, then you can construct a limit point of $a_n$ below $\alpha$ contrary to the fact that $\alpha$ is the lowest limit point.
If you consider $d<\alpha$ and $\varepsilon=\alpha-d$. So $d=\alpha-\varepsilon$. From the property above there exists $N\in\mathbb N$ such that $a_n\geq \alpha-\varepsilon=d$ for all $n\geq N$.
But this is a property for high indices $n$. In your inequality, you have just low indices. So $a_k$ can be below $d$ and so on. Therefore you need more information about $C$. Can $C$ depend on $n$ and $(a_n)_n$? Has $C$ to be positive? Or more information about $(a_n)_n$ and $d$.

Thank you so far. $C$ can not depend on $n$ but can depend on the sequence $(a_n)_n$.

If $C$ can be negative, so you could do this:
Let be $N$ like above and define $C'=\min\{a_k-d~:~k< N\}$. Then 
for $k< N$ we get
$$
a_k=a_k-d+d\geq C'+d
$$ 
and for $k\geq N$ we have $a_k\geq d$. Now we deduce
\begin{align}
\sum_{k=1}^{n-1} \frac{a_k}{k}&=\sum_{k=1}^{N-1}\frac{a_k}k+\sum_{k=N}^{n-1}\frac{a_k}{k}\geq \sum_{k=1}^{N-1}\frac{C'+d}k+\sum_{k=N}^{n-1}\frac{d}k\\&=
\sum_{k=1}^{N-1}\frac{C'}k+\sum_{k=1}^{N-1}\frac{d}k+\sum_{k=N}^{n-1}\frac{d}k\\
&=\sum_{k=1}^{N-1}\frac{C'}k+\sum_{k=1}^{n-1}\frac{d}k
\end{align}
Finally, define $C=\sum_{k=1}^{N-1}\frac{C'}k$ and you get
$$
\sum_{k=1}^{n-1}\frac{a_k}k\geq C+\sum_{k=1}^{n-1}\frac{d}k
$$
