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.
Solution 1:
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
$$\lim_{n\to\infty}\sum_{k=1}^{n} \frac{1}{k(n-k+1)}=0$$