I need to find the limit point(s) of the sequence whose general term is given by: a n

I need to find the limit point(s) of the sequence whose general term is given by:
${a}_{n}=\frac{1}{1\cdot n}+\frac{1}{2\cdot \left(n-1\right)}+\cdots +\frac{1}{n\cdot 1}$
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Aiden Norman
The sequence, ${a}_{n}$, can be written
$\begin{array}{rl}{a}_{n}& =\sum _{k=1}^{n}\frac{1}{k\left(n-k+1\right)}\\ \\ & =\frac{1}{n+1}\sum _{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k-\left(n+1\right)}\right)\\ \\ & =\frac{2}{n+1}\sum _{k=1}^{n}\frac{1}{k}\end{array}$
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
$\sum _{k=1}^{n}\frac{1}{k}\le \mathrm{log}\left(n\right)+1$
Therefore, we see that
$\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^{n}\frac{1}{k\left(n-k+1\right)}=0$