<s>[INST] Limit point of sequence whose general term is $ \frac{1}{1\cdot n} + \frac{1}{2\cdot(n-1)}+\dots+\frac{1}{n\cdot1}$ I need to find the limit point(s) of the sequence whose general term is given by:
$$a_n = \frac{1}{1\cdot n} + \frac{1}{2\cdot (n-1)}+\dots+\frac{1}{n\cdot 1}$$
My observation:

*

*$\frac{1}{1\cdot n} + \frac{1}{2\cdot (n-1)}+\dots+\frac{1}{n\cdot 1} \gt \frac{1}{n\cdot n} + \frac{1}{n\cdot (n-1)}+\dots+\frac{1}{n\cdot 1}  \gt \frac{1}{n\cdot n} + \frac{1}{n\cdot n}+\dots+\frac{1}{n\cdot n} = \frac{1}{n}$.

*The series with $a_n$ as the general term is divergent.

Intuitively I feel its a non- monotonic divergent sequence, but I am stuck at how to go further.
Kindly help. Thanks in advance.
 [/INST] The sequence, $a_n$, can be written
$$\begin{align}
a_n&=\sum_{k=1}^{n} \frac{1}{k(n-k+1)}\\\\
&=\frac1{n+1}\sum_{k=1}^{n}\left(\frac1k-\frac1{k-(n+1)}\right)\\\\
&=\frac2{n+1}\sum_{k=1}^n \frac1k 
\end{align}$$
It is easy to show that $a_n$ is monotonically decreasing and bounded below by $0$.  So $a_n$ converges.  Moreover, from the integral test, we have
$$\sum_{k=1}^n \frac1k \le \log(n)+1$$
Therefore, we see that
$$\lim_{n\to\infty}\sum_{k=1}^{n} \frac{1}{k(n-k+1)}=0$$
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