Question

What is the sum of the series $\frac{1}{1}+\left(\frac{1}{1+2}\right)+\left(\frac{1}{1+2+3}\right)+\cdots$ equal to?

Answer 1

Step 1. Find the $n$-th term of the series $\frac{1}{1}+\left(\frac{1}{1+2}\right)+\left(\frac{1}{1+2+3}\right)+\cdots$.

Consider the series $\frac{1}{1}+\left(\frac{1}{1+2}\right)+\left(\frac{1}{1+2+3}\right)+\cdots$.

Assume the series up to $n$-terms.

$\frac{1}{1}+\left(\frac{1}{1+2}\right)+\left(\frac{1}{1+2+3}\right)+\cdots +\left(\frac{1}{1+2+3+\cdots n}\right)$

Assume $S=\frac{1}{1+2+3+\cdots n}$ to determine the value of $S_n$.

$\begin{array}{rcl}{S}_n& =& \frac{1}{{\displaystyle \sum _{n=1}^n\frac{n\left(n+1\right)}{2}}}\\ & \Rightarrow & S_n=\sum _{n=1}^n\frac{2}{n\left(n+1\right)}\\ & \Rightarrow & S_n=\sum _{n=1}^n\left(\frac{2}{n}-\frac{2}{n+1}\right)\end{array}$

Hence, the $n$-th term of the series is $\begin{array}{rcl}{S}_n& =& \sum _{n=1}^n\left(\frac{2}{n}-\frac{2}{n+1}\right)\end{array}$.

Step 2. Find $S_1,S_2,S_3,\cdots ,S_n$ to calculate the sum of the series.

$\begin{array}{cl}\Rightarrow & S_2=\frac{2}{2}-\frac{2}{3}\\ \Rightarrow & S_3=\frac{2}{3}-\frac{2}{4}\\ & ⋮\\ \therefore & S_n=\frac{2}{n}-\frac{2}{n+1}\end{array}\begin{array}{}\\ \\ \\ \\ \end{array}$

Calculate the sum.

$\begin{array}{rcl}{S}_1+S_2+S_3+\dots +S_n& =& \left(\frac{2}{1}-\frac{2}{2}\right)+\left(\frac{2}{2}-\frac{2}{3}\right)+\left(\frac{2}{3}-\frac{2}{4}\right)+\cdots \left(\frac{2}{n}-\frac{2}{n+1}\right)\\ & \Rightarrow & S_1+S_2+S_3+\dots +S_n=\frac{2}{1}-\frac{2}{n+1}\\ & \Rightarrow & S_1+S_2+S_3+\dots +S_n=\frac{2\left(n+1\right)-2}{n+1}\\ \therefore S_1+S_2+S_3+\dots +S_n& =& \frac{2n}{n+1}\end{array}$

Hence, the sum of the given series is $\frac{2n}{n+1}$.