Infinite Series

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Hi, and welcome to this video about infinite series! In this video, we will explore what an infinite series is, convergence vs. divergence, properties of convergent series, and the nth term test for divergence. Let’s get started!

So first, what is an infinite series? An infinite series is an infinite sum, which looks like this:

\(a_{1} + a_{2} + a_{3} + a_{4} + … + a_{n} + …\)

The a’s in the expression are called terms and can be pretty much anything—numbers, functions, etc.—and the subscripts represent the term number, kind of like a counter. In this series, “a sub 1” is the first term of the series, “a sub 2” is the second term, and so on.

When writing out equations, sigma notation is typically used to represent infinite sums. Instead of the expression we just saw, we can use a capital Greek sigma to write the summation of n equals 1, to infinity, of a sub n \(\sum_{n=1}^{\infty}a_{n}\) or simply the summation of a sub n \(\sum a\) when the series is infinite.

Suppose we had the series the summation of n equals 1 to infinity, of 2n plus 1 \(\sum_{n=1}^{\infty}2n+1\). This is what the first five terms written out would look like: [image onscreen at 1:08}

Usually, with infinite series, we are not interested in just a few terms – we are interested in all of them.

Now, let’s talk about convergence and divergence. If we were able to add up all of the terms in an infinite series, we would find that the addition of more and more terms would either “settle” on a value or keep growing.

This series, the sum of n equals 1 to infinity, of 1 over 3, to the n minus 1, equals 1 over 1, plus 1 over 3, plus 1 over 9, plus 1 over 27, and so on.
And it ends up settling on the value 3 over 2.

This series \(\sum_{n=1}^{\infty}\frac{1}{3^{n-1}} = 1/1 + 1/3 + 1/9 + 1/27 + …\) ends up settling on the value \(\frac{3}{2}\). In other words, the series is convergent and it converges on the value 3 over 2. Notice how, when n gets large enough, adding 1 over 3, to the n minus 1, \(\frac{1}{3^{n-1}}\)won’t really affect the sum, because each successive fraction will be smaller than the one before.

On the other hand, our first example series \(\sum_{n=1}^{\infty}2n+1 = 3 + 5 + 7 + 9 + 11\) just keeps growing and is divergent. Each term is larger than the last, causing the sum to keep getting larger and diverge.

To put it simply, convergent series have a limit; divergent series do not.

Once you have determined that two series are convergent, there are 3 important properties to remember:

1. The series of a sum is the sum of the series:

The summation of, a sub n plus b sub n, is equal to the summation of a sub n, plus the summation of b sub n.

\(\sum(a_{n}+b_{n})=\sum a_{n}+\sum b_{n}\)

2. The series of a difference is the difference of the series

The summation of, a sub n minus b sub n, is equal to the summation of a sub n, minus the summation of b sub n.

\(\sum(a_{n}-b_{n})=\sum a_{n}-\sum b_{n}\)

3. The series of a multiple is the multiple of the series

The summation of k, a scalar, times a sub n, is equal to k times the summation of a sub n.

\(\sum ka_{n}=k\sum a_{n}\)

Here are two convergent series to illustrate: \(\sum\frac{1}{n^{2}}=\frac{\pi^{2}}{6}\) and \(\sum\frac{1}{n^{n}}\approx 1.29\)

[onscreen at 2:56]

Now let’s shift to divergence. One of the first tests to perform on an infinite series is the nth term test for divergence. The advantage of this test is that if a series is divergent, you can stop there. The disadvantage is that knowing a series is not divergent by this test is not enough to automatically call it convergent, so the test becomes inconclusive.

How does this test work? Recall, we said that a series is convergent if it has a limit and divergent if it does not. All we need to do is take the limit of the nth term of the sequence. If the limit is not equal to 0, the series is divergent. If the limit is equal to 0, our result is inconclusive.

Back to our first example. We know that the sum from n equal 1 to infinity, of 2n plus 1, \(\sum_{n=1}^{\infty} 2n=1\) is divergent. The test confirms this.

\(\lim_{n\rightarrow \infty }\frac{1}{n^{2}}=\infty\)

Since this limit is not equal to 0, the series is divergent.

We know that the sum of n equals 1 to infinity, of 1 over n squared, \(\sum_{n=1}^{\infty}\frac{1}{n^{2}}\) is convergent, but this test alone doesn’t give us much to go on:

\(\lim_{n\rightarrow \infty }\frac{1}{n^{2}}=0\)

This is inconclusive because the limit is equal to zero. As mentioned before, there are other more involved tests to determine whether series converge or diverge, but we’ll cover that in another video.

I hope this video helped you understand infinite series and how they work! Thanks for watching, and happy studying!

Infinite Series

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