Find the value of $\lim_{n\rightarrow\infty}\frac{a_n}{1\cdot 2}+\frac{a_{n-1}}{2\cdot 3}+\ldots +\frac{a_1}{n(n+1)}$ $\displaystyle \lim_{n\rightarrow\infty}\frac{a_n}{1\cdot 2}+\frac{a_{n-1}}{2\cdot 3}+\ldots +\frac{a_1}{n(n+1)}$ if $\lim_{n\rightarrow\infty}a_n=a$
Note that $\displaystyle\sum_{k=1}^n\frac{a_{n-k+1}}{n^2}\le S_n\le \sum_{k=1}^n\frac{a_{n-k+1}}{k^2}$
Can I bound $S_n$ by $a$ instead of $a_{n-k+1}$ in the numerator instead?
 A: We will show that the limit is $a$.
Let $\varepsilon>0$. First, $(a_n)$ is bounded, so $|a_n|<M$ for some $M>0$, for all $n\in\mathbb N$. There exists $n_0\in\mathbb N$ such that, for all $n\geq n_0$, $|a_n-a|<\frac{\varepsilon}{6}$. Also, there exists $n_1\in\mathbb N$ with $n_1>n_0$ such that, if $n\geq n_1$, then $\sum_{k=n_1}^{\infty}\frac{1}{k(k+1)}<\frac{\varepsilon}{6M}$. Finally, there exists $n_2\in\mathbb N$ such that, if $n\geq n_2$, we have that $\frac{1}{n+1}<\frac{\varepsilon}{3M}$. Then, if $n\geq n_0,2n_1,n_2$,$$\left|\frac{a_n}{1\cdot2}+\frac{a_{n-1}}{2\cdot3}+\dots+\frac{a_1}{n(n+1)}-a\right|=$$$$\left|\frac{a_n-a}{1\cdot2}+\frac{a_{n-1}-a}{2\cdot3}+\dots+\frac{a_1-a}{n(n+1)}-a+\frac{a}{1\cdot2}+\frac{a}{2\cdot3}+\dots+\frac{a}{n(n+1)}\right|\leq$$$$\left|\frac{a_n-a}{1\cdot2}+\frac{a_{n-1}-a}{2\cdot3}+\dots+\frac{a_1-a}{n(n+1)}\right|+\left|\frac{a}{1\cdot2}+\frac{a}{2\cdot3}+\dots+\frac{a}{n(n+1)}-a\right|.$$ The term in the second absolute value is equal to $$\left|a-\frac{a}{2}+\frac{a}{2}-\frac{a}{3}+\dots+\frac{a}{n}-\frac{a}{n+1}-a\right|=\left|-\frac{a}{n+1}\right|\leq\frac{M}{n+1}<\frac{\varepsilon}{3},$$ since $n\geq n_2$. For the first term, you have that $$\left|\frac{a_n-a}{1\cdot2}+\frac{a_{n-1}-a}{2\cdot3}+\dots+\frac{a_1-a}{n(n+1)}\right|\leq$$$$\left|\frac{a_n-a}{1\cdot2}+\frac{a_{n-1}-a}{2\cdot3}+\dots+\frac{a_{n_1}-a}{(n-n_1+1)(n-n_1+2)}\right|+$$$$\left|\frac{a_{n_1-1}-a}{(n-n_1+2)(n-n_1+3)}+\dots+\frac{a_1-a}{n(n+1)}\right|.$$ Since $n_1>n_0$, the first term here is bounded by $$\left|\frac{\varepsilon/6}{1\cdot2}+\dots+\frac{\varepsilon/6}{(n-n_1+1)(n-n_1+2)}\right|<\frac{2\varepsilon}{6}=\frac{\varepsilon}{3},$$ and the second term is bounded by $$2M\sum_{k=n-n_1+2}^\infty\frac{1}{k(k+1)}<\frac{2M\varepsilon}{6M}=\frac{\varepsilon}{3},$$ since $n-n_1+2\geq n_1$. After adding those inequalities, we get that this limit is equal to $a$.
