Solve the recursion $a_{n+1} = a_n\left(1+\frac{1}{n}\right) +1; a_1=1$
Can this recursion be solved? I don't think generating functions would help. I tried converting it to a telescoping series as well and ended up with $$a_{n+1} = \sum_{k=1}^{n} \left\{a_k \left( \frac{1}{k} \right)\right\} + (n+1)$$
A bit of observations gave me a hint that $\exists n$ such that $a_n > x ~ \forall x\in\mathbb{R}$. Is that something correct? I am not sure but can that be proven? (using bounding?)
$$ n a_{n+1} = a_n (n+1) + n \tag{1}$$ can be rearranged as $$ \frac{a_{n+1}}{n+1}-\frac{a_n}{n} = \frac{1}{n+1}\tag{2} $$ hence by summing both sides of $(2)$ over $n\in[1,N]$ we get $$ \frac{a_{N+1}}{N+1}-1 = \frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{N+1} = H_{N+1}-1 \tag{3} $$ then $$ a_N = N H_N = N\sum_{k=1}^{N}\frac{1}{k}.\tag{4} $$