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Calculate the first eight terms of the sequence of partial sums correct to four decimal places. Does it appear that the series is convergent or divergent?$$\sum_{n=1}^{\infty} \frac{1}{n^{3}}$$

Convergent

Calculus 2 / BC

Chapter 8

SERIES

Section 2

Series

Sequences

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welcome number on My Name is Kevin Chirac. Let's take a look at the 1st 8 terms of an infinite Siri's that is denoted as follows. It'll be n equals one to infinity of one over n Cubed. And let's look at the 1st 8 1st 8 terms in this, uh, the Siri's. So now we're looking for 1/1 cubed, plus one over two cubed, plus 1/3, cubed all the way up until we get to 1/8. Cute, but these are simply the terms inthe e sequence. What we'd like to do is look at the partial summation. So let's now look at F sub one, which would be the partial summation up till the first term. And so that's just something to be won over one Cubed, which is one let's look at as to well as To is going to be the partial summation from the first term until the second term. So now we have one plus 1/8 that's going to be 1.1 to 5 for s three. It would be up until and including the third term, So now we've got 1.125 plus 1 27th So let's take a moment and pull out a calculators. And if we type into our calculators one 0.1 to 5 plus 1 27th and get a total of 1.16 to 035 decibels So one point 16 203 and that would be s three. We could continue this process all the way up until we get Teoh, including that eighth term, and we will be wanting to look at whether or not this sequence converges. Now. One thing that I will tell you is that this is going to be a sequence that converges, and the reason we can tell is because there's going to be more and more of these decimal places that are going to begin toe walk. In other words, they're not going to appear to grow even as we add more terms. So the number is going to approach something. But these individual digits are going to begin Teoh that the change will be so small that are larger digits will remain unaffected. So I hope

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