Original subtitles

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Today we're going to be talking about how to calculate the first several terms in a sequence of partial

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sums.

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And in this particular problem we've been asked to calculate the first several terms of this sequence

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and divide by 1 plus the square root of N from N equals 1 to infinity.

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Now before we talk about how to calculate the first terms I want to as a reminder make sure that we

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understand the difference between a sequence and a sequence of partial sums.

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So in a regular series here we might have the series a Sabine.

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Right.

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That's different than how we denote the series of partial sums.

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Es seben So a sideband is our typical series as Serbin is our series or a sequence of partial sums.

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Now when it comes to a regular series we might have some regular series like for example one two three

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four five right.

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The series of partial sums will be each one of these terms added together with the terms before it.

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So the first terms will be the same right one in one.

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But then the second term I'm going to add two to the first term in my series of partial sum so one plus

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two gives me three.

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Then I add the next term to that three plus three gives me six to six I add four and I get 10 to 10

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I add 5 and I get 15 and I could keep going.

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So you can see how these series are different.

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We have the regular series a sidebar in the series of partial sums.

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We're taking partial sums of this series.

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In other words the sum of the series up to the given point so 10 is the sum of the series up until the

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fourth term of the series.

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So that's why we call it the series of partial sums.

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So in order to calculate the first terms in the sequence of partial sums here what we want to do is

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just start playing in values of n.

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Based on whatever we're given here so we were given in this particular problem and equals 1.

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So we start with an equals 1.

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So we say at any equals 1 we plug in 1 to our series here and we get 1 divided by 1 plus the square

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root of 1.

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And of course when we evaluate that we'll get the squared of one to be one one plus one is two.

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And so we get a value of 1 1/2 which is equal to zero point five.

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And we can just add a couple extra zeros because we know we're going to need them later.

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OK.

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So then we plug in a value and equals two and we can just keep doing this right we'll get 2 divided

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by 1 plus the square root of two.

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And if we evaluate that we'll get approximately point eight to a four that's an approximate value.

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But remember because we're dealing with the sequence of partial sums we have to add this value the value

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of the second term to the previous value that we got.

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So we have to add that two point five point eight to eight four plus point five is going to give us

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the value of the second term in a sequence of partial sums which is approximately 1 point 3 2 8 4 and

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we can round that to about 3 to 8.

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So let's just do one more term we'll get an equals three we get three over 1 plus the square root of

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3.

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When we evaluate that on our calculator we'll get approximately 1 point 0 9 8 or so.

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But remember that's only the third term of a seven in order to get the third term of 7.

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We need to add it to our previous value of 1.3 to 8 when we do that and we round we get approximately

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two point four to seven for the third term in our sequence of partial sum.

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So remember that value we get here when we just plug in our value for end directly is the value of a

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7 when we add it to the previous term.

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We get the value in the sequence of partial sums as Subban.

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So here's how we generate the values of S-band and we can just keep going.

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And he calls for unequals 5 and 6 and we just added to that previous sum that we found what we'd see

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is that we get approximately three point 6 0 for any calls four for N equals five.

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We get approximately five point three zero five for any six.

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We get approximately seven point zero for four and we keep going.

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One thing we can say just looking at this sequence of partial sums though is that the sequence appears

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to be divergent because this value just keeps getting larger and larger and larger.

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And in fact the change between each term keeps getting larger and larger.

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This sequence of partial sums as then appears to be divergent or just appears divergent and that's one

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conclusion that we can try to draw from the list of terms that we can it here.