Limit of the series vs. sum of the series

 
 
Limit vs. sum of the series blog post.jpeg
 
 
 

The limit of the series vs. the sum of the series

Sometimes it’s easy to forget that there’s a difference between the limit of an infinite series and the sum of an infinite series.

The limit of a series is the value the series’ terms are approaching as ???n\to\infty???.

The sum of a series is the value of all the series’ terms added together.

They’re two very different things, and we use a different calculation to find each one. Let’s find both the limit and the sum of the same series so that we can see the difference.

Krista King Math.jpg

Hi! I'm krista.

I create online courses to help you rock your math class. Read more.

 
 
 

Comparing the limit and sum of the series


 
Krista King Math Signup.png
 
Calculus 2 course.png

Take the course

Want to learn more about Calculus 2? I have a step-by-step course for that. :)

 
 

 
 

How to find the limit of the series and sum of the series for the same series

Example

Find the limit and the sum of the series.

???\sum^{\infty}_{n=1}\frac{2^{3n}}{64^{\frac{n}{2}}}???

To find the limit of the series, we’ll identify the series as ???a_n???, and then take the limit of ???a_n??? as ???n\to\infty???.

???\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{2^{3n}}{64^{\frac{n}{2}}}???

???\lim_{n\to\infty}\frac{\left(2^{3}\right)^n}{\left(64^{\frac{1}{2}}\right)^n}???

???\lim_{n\to\infty}\frac{8^n}{\left(\sqrt{64}\right)^n}???

???\lim_{n\to\infty}\frac{8^n}{8^n}???

???\lim_{n\to\infty}1???

???1???

The limit of the series is ???1???.

Limit vs. sum of the series for Calculus 2.jpg

They’re two very different things, and we use a different calculation to find each one.

To find the sum of the series, we’ll expand the series.

???\sum^{\infty}_{n=1}\frac{2^{3n}}{64^{\frac{n}{2}}}=\frac{2^{3(1)}}{64^{\frac{1}{2}}}+\frac{2^{3(2)}}{64^{\frac{2}{2}}}+\frac{2^{3(3)}}{64^{\frac{3}{2}}}+...???

???\sum^{\infty}_{n=1}\frac{2^{3n}}{64^{\frac{n}{2}}}=\frac{2^3}{\sqrt{64}}+\frac{2^6}{\left(\sqrt{64}\right)^2}+\frac{2^9}{\left(\sqrt{64}\right)^3}+...???

???\sum^{\infty}_{n=1}\frac{2^{3n}}{64^{\frac{n}{2}}}=\frac{8}{8}+\frac{64}{8^2}+\frac{512}{8^3}+...???

???\sum^{\infty}_{n=1}\frac{2^{3n}}{64^{\frac{n}{2}}}=\frac{8}{8}+\frac{64}{64}+\frac{512}{512}+...???

???\sum^{\infty}_{n=1}\frac{2^{3n}}{64^{\frac{n}{2}}}=1+1+1+...???

???\sum^{\infty}_{n=1}\frac{2^{3n}}{64^{\frac{n}{2}}}=\infty???

Every term in our series will be equal to ???1???. Since we have an infinite number of terms in our series, we can say that the sum is infinite.

We can see that the limit of the series is ???1???, but the sum of the same series is ???\infty???. 

 
Krista King.png
 

Get access to the complete Calculus 2 course