Find the sum of the series $\frac{1}{2\cdot 3}2+\frac{2}{3\cdot 4}2^2+\frac{3}{4\cdot 5}2^3+\cdots$ up to $n$ terms

Question

Find the sum of the following series up to $n$ terms

$$\frac{1}{2\cdot 3}2+\frac{2}{3\cdot 4}2^2+\frac{3}{4\cdot 5}2^3+\cdots$$

Can someone please give any clue?

Answer 1

Rewrite the sum \begin{eqnarray*} \sum_{i=1}^{n} \frac{i}{(i+1)(i+2)} 2^{i} \end{eqnarray*} Now do partial fractions \begin{eqnarray*} \frac{i}{(i+1)(i+2)} =\frac{2}{n+2} - \frac{1}{n+1}. \end{eqnarray*} Now not that the powers of $2$ cause this sum to be telescoping, so Rewrite the sum \begin{eqnarray*} \sum_{i=1}^{n} \frac{i}{(i+1)(i+2)} 2^{i}=\frac{2^{n+1}}{n+2}-1. \end{eqnarray*}

Answer 2

Usiing telescopic sum $$\sum_{n\geq1}^k\dfrac{n}{(n+1)(n+2)}2^n=\sum_{n\geq1}^k\left(\dfrac{2^{n+1}}{n+2}-\dfrac{2^{n}}{n+1}\right)=\dfrac{2^{k+1}}{k+2}-1$$

Answer 3

You are interested in the sum $$ S_{N}\equiv\sum_{n=1}^{N}\frac{n}{\left(n+1\right)\left(n+2\right)}2^{n}. $$ As Donald suggests, you can use the partial fraction expansion $$ \frac{n}{\left(n+1\right)\left(n+2\right)}=\frac{2}{n+2}-\frac{1}{n+1} $$ to get \begin{multline*} S_{N}=\sum_{n=1}^{N}\frac{2}{n+2}2^{n}-\sum_{n=1}^{N}\frac{1}{n+1}2^{n}=\sum_{n=1}^{N}\frac{1}{\left(n+1\right)+1}2^{n+1}-\sum_{n=1}^{N}\frac{1}{n+1}2^{n} =\sum_{n=2}^{N+1}\frac{1}{n+1}2^{n}-\sum_{n=1}^{N}\frac{1}{n+1}2^{n}\\=\sum_{n=2}^{N+1}\frac{1}{n+1}2^{n}-\sum_{n=1}^{N}\frac{1}{n+1}2^{n}=\frac{1}{N+2}2^{N+1}-\frac{1}{2}2^{1} =\frac{1}{N+2}2^{N+1}-1. \end{multline*} From the above, it is clear that $S_{N}\rightarrow\infty$ as $N\rightarrow\infty$.