Sequences & Series – Problem 6: Find the general formula of

Let $S_n$ be the sum of the first $n$ terms of the sequence $\{a_n\}$. Suppose $$S_n=\frac{n+1}{2}a_n,$$ and $a_1=1$.

(1) Find the general formula of $\{a_n\}$.

(2) If $$b_n=\frac{1}{a_na_{n+2}},$$ find $T_n$, the sum of the first $n$ terms of $\{b_n\}$.

(1) When $n\ge2$, $$a_n=S_n-S_{n-1}=\frac{n+1}{2}a_n-\frac{n}{2}a_{n-1}.$$ Rearranging gives $$na_{n-1}=(n-1)a_n\Rightarrow \frac{a_{n-1}}{n-1}=\frac{a_n}{n}\ (n\ge2).$$ Since $a_1=1$, we have $\dfrac{a_n}{n}=1$, hence $a_n=n$.

(2) $$b_n=\frac1{n(n+2)}=\frac12\left(\frac1n-\frac1{n+2}\right).$$ Therefore $$T_n=\sum_{k=1}^n b_k =\frac12\sum_{k=1}^n\left(\frac1k-\frac1{k+2}\right) =\frac12\left(1+\frac12-\frac1{n+1}-\frac1{n+2}\right).$$ So $$T_n=\frac34-\frac12\left(\frac1{n+1}+\frac1{n+2}\right).$$

(1) The sequence satisfies $a_n=n$. This follows from the recurrence obtained by subtracting $S_{n-1}$ from $S_n$, which yields the invariant ratio $\dfrac{a_n}{n}=1$.

(2) After substituting $a_n=n$, we get $b_n=\dfrac1{n(n+2)}$, which is telescoping after decomposition. Thus $$T_n=\frac34-\frac12\left(\frac1{n+1}+\frac1{n+2}\right).$$ This gives the exact partial-sum formula for $\{b_n\}$.