Science:Infinite Series Module/Units/Unit 1/1.2 Sigma Notation/1.2.05 Changing Summation Limits Example

Problem

Change the following summation

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \sum_{k=3}^{\infty} \frac{k}{2+k} = \frac{3}{2+3}+\frac{4}{2+4}+\frac{5}{2+5}+\ldots , \quad (1)\end{align} }

so that the index of summation start at 1 instead of at 3.

Complete Solution

Although not necessary, we will use two method for solving this problem: Method 1 and Method 2.

Method 1

Introducing the transformation

$\begin{aligned} j = k - 2, \end{aligned}$

so that when $k = 3, j = 1$ gives us

$\sum_{k = 3}^{\infty} \frac{k}{2 + k} = \sum_{j = 1}^{\infty} \frac{j + 2}{2 + (j + 2)} = \frac{3}{2 + 3} + \frac{4}{2 + 4} + \frac{5}{2 + 5} + \dots$

as desired.

Method 2

Our procedure is to add and subtract terms in the sum to shift our index to 1:

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \sum_{k=3}^{\infty} \frac{k}{2+k} &= \frac{3}{2+3}+\frac{4}{2+4}+\frac{5}{2+5}+\ldots && (2) \\ \\ &= \Big( \mathbf{\color{Purple}{ \frac{2}{2+2} }} - \frac{2}{2+2} \Big) + \frac{3}{2+3}+\frac{4}{2+4}+\frac{5}{2+5}+\ldots && (3) \\ \\ &= \Big( - \frac{2}{2+2} \Big) + \mathbf{\color{Purple}{ \frac{2}{2+2} }} + \frac{3}{2+3}+\frac{4}{2+4}+\frac{5}{2+5}+\ldots && (4) \\ \\ &= \Big( \mathbf{\color{Mahogany}{ \frac{1}{2+1} }} - \frac{1}{2+1} \Big) + \frac{1}{2+1} + \mathbf{\color{Purple}{ \frac{2}{2+2} }} + \frac{3}{2+3}+\frac{4}{2+4}+\frac{5}{2+5}+\ldots && (5) \\ \\ &= \Big( - \frac{1}{2+1} \Big) + \mathbf{\color{Mahogany}{ \frac{1}{2+1} }} +\frac{2}{2+2} + \frac{3}{2+3}+\frac{4}{2+4}+\frac{5}{2+5}+\ldots && (6) \\ \\ &= \Big( - \frac{1}{2+1} - \frac{2}{2+2} \Big) + \sum_{k=1}^{\infty} \frac{k}{2+k} && (7) \\ \\ &= - \frac{5}{6} + \sum_{k=1}^{\infty} \frac{k}{2+k} && (8) \\ \\ \end{align}}

as desired. Therefore

$\sum_{k = 3}^{\infty} \frac{k}{2 + k} = - \frac{5}{6} + \sum_{k = 1}^{\infty} \frac{k}{2 + k}$

Discussion of Some Steps

Method 1

The transformation $j = k - 2$ , was chosen to that the index $j$ would start at 1.

Method 2

Most steps in this approach involved straightforward algebraic manipulation. Steps (3) and (5) involve adding and subtracting terms in a way that will allow us to change the summation limits.

More precicesly, in Step (3) we added and substracted the $k = 2$ term, and in Step (5) we added and subtracted the $k = 1$ term.

Potential Challenges

Method 2 Requires More Work, So Why Should I Use It?

You may not have a choice. In some circumstances, you need to convert your sum in a way that does not change the general term.