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

## Problem

Change the following summation

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

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

so that when ${\displaystyle k=3,\ j=1}$ gives us

${\displaystyle \sum _{k=3}^{\infty }{\frac {k}{2+k}}=\sum _{j=1}^{\infty }{\frac {j+2}{2+(j+2)}}={\frac {3}{2+3}}+{\frac {4}{2+4}}+{\frac {5}{2+5}}+\ldots }$

as desired.

### Method 2

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

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

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

## Discussion of Some Steps

### Method 1

The transformation ${\displaystyle j=k-2}$, was chosen to that the index ${\displaystyle j}$ 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 ${\displaystyle k=2}$ term, and in Step (5) we added and subtracted the ${\displaystyle k=1}$ 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.