# Finding sum of infinite series: sums of two series together

• Ryaners
In summary, an infinite series is a sum of an infinite number of terms that follow a specific pattern. To find the sum of an infinite series, one must determine if the series converges or diverges and use a specific formula or method accordingly. A finite series has a limited number of terms and a definite sum, while an infinite series has an unlimited number of terms and may or may not have a definite sum. Not all infinite series can be summed, only convergent series can have a finite sum. Infinite series have various real-world applications in fields such as physics, engineering, finance, economics, and art.
Ryaners

## Homework Statement

Find the sum of the following series:
$$\left( \frac 1 2 + \frac 1 4 \right) + \left( \frac 1 {2^2} + \frac 1 {4^2} \right) +~...~+ \left( \frac 1 {2^k} + \frac 1 {4^k} \right) +~...$$

## Homework Equations

$$\sum_{n = 1}^{\infty} \left( u_k+v_k \right) = \sum_{n = 1}^{\infty} u_k + \sum_{n = 1}^{\infty} v_k$$
$$\sum_{n = 1}^{\infty} cu_k = c \sum_{n = 1}^{\infty} u_k$$
Sum of a geometric series ## \sum_{n = 1}^{\infty} ar^k = \frac a {1-r} ##

## The Attempt at a Solution

The question referred back to the two relations I've put in the 'relevant equations' section - I assumed I'm meant to use the first one. So I split the sum into two sums as follows:
$$\sum_{n = 1}^{\infty} \left( \frac 1 {2^k} + \frac 1 {4^k} \right) = \sum_{n = 1}^{\infty} \frac 1 {2^k} + \sum_{n = 1}^{\infty} \frac 1 {4^k}$$
For the left term, I got:
$$\sum_{n = 1}^{\infty} \frac 1 {2^k} = \sum_{n = 1}^{\infty} \left( \frac 1 2 \right)^k = \frac 1 {1- \frac 1 2} = 2$$
Similarly for the right term:
$$\sum_{n = 1}^{\infty} \frac 1 {4^k} = \sum_{n = 1}^{\infty} \left( \frac 1 4 \right)^k = \frac 1 {1- \frac 1 4} = \frac 4 3$$
Then adding these together gives me ## 3 \frac 1 3 ##, which is... the wrong answer

I can see that ## \frac 1 {4^k} = \frac 1 {2^{2k}} ## - has that got something to do with the solution?

Edit: I should say that the answer in the back of the book is ## \frac 4 3 ## which makes me think that there must be a way that the first series is 'incorporated into' the second, but I can't quite get my head around that...!

Ryaners said:

## Homework Statement

Find the sum of the following series:
$$\left( \frac 1 2 + \frac 1 4 \right) + \left( \frac 1 {2^2} + \frac 1 {4^2} \right) +~...~+ \left( \frac 1 {2^k} + \frac 1 {4^k} \right) +~...$$

## Homework Equations

$$\sum_{n = 1}^{\infty} \left( u_k+v_k \right) = \sum_{n = 1}^{\infty} u_k + \sum_{n = 1}^{\infty} v_k$$
$$\sum_{n = 1}^{\infty} cu_k = c \sum_{n = 1}^{\infty} u_k$$
Sum of a geometric series ## \sum_{n = 1}^{\infty} ar^k = \frac a {1-r} ##
The formula is valid if the sum goes from k=0

Ryaners said:
...

## Homework Equations

[/B]$$\sum_{n = 1}^{\infty} \left( u_k+v_k \right) = \sum_{n = 1}^{\infty} u_k + \sum_{n = 1}^{\infty} v_k$$ $$\sum_{n = 1}^{\infty} cu_k = c \sum_{n = 1}^{\infty} u_k$$ Sum of a geometric series ## \sum_{n = 1}^{\infty} ar^k = \frac a {1-r} ##
A couple of problems here.

First, the index in the Sigma notation should match the index in the items you are summing.

Then, as @ehild has pointed out, your index should start at zero, not 1 (unless you adjust for that).

Sum of a geometric series:
##\displaystyle \sum_{k = 0}^{\infty} ar^k = \frac a {1-r} ##
or​
##\displaystyle \sum_{k = 1}^{\infty} ar^k = \frac {ar} {1-r} ##

$$\sum_{n = 1}^{\infty} \frac 1 {2^k} = \sum_{n = 1}^{\infty} \left( \frac 1 2 \right)^k = \frac {\color{green}{1}} {1- \frac 1 2} = 2$$
The first term(a) is not 1.

Buffu said:
$$\sum_{n = 1}^{\infty} \frac 1 {2^k} = \sum_{n = 1}^{\infty} \left( \frac 1 2 \right)^k = \frac {\color{green}{1}} {1- \frac 1 2} = 2$$
The first term(a) is not 1.

$$\sum_{n = 1}^{\infty} \frac 1 {2^k} = 2 \; \Longleftarrow \; \text{FALSE!}$$
$$\begin{array}{lcc} \frac{1}{2}+\frac{1}{4} &=& \frac{3}{4}\\ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} &=& \frac{7}{8}\\ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +\frac{1}{16} &=& \frac{15}{16}\\ \vdots&\vdots&\vdots \end{array}$$
The sums are creeping up to 1, not to 2.

Ah ok, my mistake. Thank you all for pointing that out, I didn't realize the formula would change if the index changed, though of course that makes sense. Cheers!

I wouldn't consider it solved until you give a complete solution.

Yes, it's likely that now you do know how to solve this.​

Ray Vickson said:
$$\sum_{n = 1}^{\infty} \frac 1 {2^k} = 2 \; \Longleftarrow \; \text{FALSE!}$$
$$\begin{array}{lcc} \frac{1}{2}+\frac{1}{4} &=& \frac{3}{4}\\ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} &=& \frac{7}{8}\\ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +\frac{1}{16} &=& \frac{15}{16}\\ \vdots&\vdots&\vdots \end{array}$$
The sums are creeping up to 1, not to 2.
Yes I know, that is the error in the post.
He forgot to account for first term of the GP (a) in the formula.

I should have wrote the formula in quote box.

SammyS said:
I wouldn't consider it solved until you give a complete solution.

Yes, it's likely that now you do know how to solve this.​

So by multiplying the formula by ##r## in order to start the index at 1 instead of 0, it becomes ## \sum_{n=1}^{\infty} \frac {ar} {1-r} ## where ##a = a_0 = 1## for both series (because when summing a geometric series from ##n=1##, the first term is ##a_1 = ar##). The sum of each series is therefore 1 and ##\frac 1 3## respectively giving a total of ##\frac 4 3##.

Buffu said:
Yes I know, that is the error in the post.
He forgot to account for first term of the GP (a) in the formula.

* she :)

Thanks again folks!

SammyS
@Ryaners, please post questions about infinite series in the Calc & Beyond section. I have moved this thread.

## 1. What is an infinite series?

An infinite series is a sum of an infinite number of terms that follow a specific pattern. Each term in the series is added together to create a sum, and as more terms are added, the sum either approaches a specific value or diverges.

## 2. How do you find the sum of an infinite series?

To find the sum of an infinite series, you must first determine if the series converges or diverges. If it converges, you can use a specific formula or method to find the sum, such as the geometric series formula or the telescoping series method. If the series diverges, then there is no finite sum and the series is said to have no sum.

## 3. What is the difference between a finite and an infinite series?

A finite series has a limited number of terms and a definite sum, while an infinite series has an unlimited number of terms and may or may not have a definite sum. In other words, a finite series will eventually end and have a final value, while an infinite series will continue on forever or approach a specific value.

## 4. Can any infinite series be summed?

No, not all infinite series can be summed. Only convergent series, or series that approach a specific value, can have a finite sum. Divergent series, or series that do not approach a specific value, cannot have a finite sum.

## 5. What are some real-world applications of infinite series?

Infinite series are commonly used in physics, engineering, and other scientific fields to model and solve real-world problems. They can also be used in finance and economics to calculate compound interest and other financial growth patterns. Additionally, infinite series have been used in art and music to create fractal patterns and intricate melodies.

Replies
3
Views
941
Replies
1
Views
640
Replies
3
Views
547
Replies
3
Views
783
Replies
2
Views
1K
Replies
1
Views
774
Replies
1
Views
692
Replies
1
Views
908
Replies
2
Views
1K
Replies
8
Views
1K