Sum of the infinite series The series is $$\frac{5}{1\cdot2}\cdot\frac{1}{3}+\frac{7}{2\cdot3}\cdot\frac{1}{3^2}+\frac{9}{3\cdot4}\cdot\frac{1}{3^3}+\frac{11}{4\cdot5}\cdot\frac{1}{3^4}+\cdots$$
This is my attempt:
$$T_n=\frac{2n+3}{n(n+1)}\cdot\frac{1}{3^n}$$
Assuming $$\frac{2n+3}{n(n+1)}=\frac{A}{n}+\frac{B}{n+1}$$
we find $A=3,B=-1.$ Putting these values in $T_n$ we get,
$$T_n=\frac{1}{n}\cdot\frac{1}{3^{n-1}}-\frac{1}{n+1}\cdot\frac{1}{3^{n}}$$
How do I find the sum of the series from here $?$
 A: Just try to write the expression for $T_{n+1}$ and note that the series telescopes. 
$$T_n=\frac{1}{n}\cdot\frac{1}{3^{n-1}}-\color{purple}{\frac{1}{n+1}\cdot\frac{1}{3^{n}}} \\
T_{n+1}=\color{purple}{\frac{1}{n+1}\cdot\frac{1}{3^{n}}}-\frac{1}{n+2}\cdot\frac{1}{3^{n+1}}$$
We see the terms are cancelling, so 
$$\begin{align}
\sum_{i=1}^{n} T_i &= \frac{1}{1}\frac{1}{3^{0}}
-\color{blue}{\frac{1}{2}\frac{1}{3^{1}}+\frac{1}{2}\frac{1}{3^{1}}}...
-\color{red}{\frac{1}{n-1}\frac{1}{3^{n-2}}+\frac{1}{n-1}\frac{1}{3^{n-2}}} 
-\frac{1}{n}\frac{1}{3^{n-1}}\\
 &= 1-\frac{1}{n3^{n-1}}
\end{align}$$
Therefore as $n\to \infty$, we see the sum tends to $1$.
A: $\sum_{i=1}^{n} T_i =S_n - R_n$ , where 
$S_n := \sum_{i=1}^{n} \dfrac{1}{i}\dfrac{1}{3^{i-1}};$
$R_n :=\sum_{i=1}^{n} \dfrac{1}{i+1} \dfrac{1}{3^i}.$
Change the dummy index  in $R_n:$
$k= i +1$, then
$R_n=\sum_{k=2}^{n+1} \dfrac{1}{k}\dfrac{1}{3^{k-1}} =$
$(1 + \sum_{k=2}^{n}\dfrac{1}{k}\dfrac{1}{3^{k-1}}) -1 + 
\dfrac{1}{n+1}\dfrac{1}{3^n}$
$= S_n - 1 + \dfrac{1}{n+1}\dfrac{1}{3^n}.$
Hence:
$T_n = S_n - R_n= 1 - \dfrac{1}{ n+1}\dfrac{1}{3^n}.$
A: $$Sum=\sum_{n=1}^{\infty} \frac{1}{n}.\frac{1}{3^{n-1}}-\frac{1}{n+1}.\frac{1}{3^{n}}$$
$$Sum=\sum_{n=1}^{\infty} \frac{1}{n}.\frac{1}{3^{n-1}}-\sum_{n=1}^{\infty} \frac{1}{n+1}.\frac{1}{3^{n}}$$
$$Sum=\sum_{n=1}^{\infty} \frac{1}{n}.\frac{1}{3^{n-1}}-\sum_{n=2}^{\infty} \frac{1}{n}.\frac{1}{3^{n-1}}$$
$$Sum=\sum_{n=1}^{1} \frac{1}{n}.\frac{1}{3^{n-1}}=1$$
