Original subtitles

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In this video we're talking about the difference between the limit of a series and the sum of a series.

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And in this particular problem we have the infinite series from equals one to infinity of 4000 96 race

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to the end over to power divided by a race to the two and power.

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So when we're taking the limit of an infinite series we're interested in the value that the series approaches

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as and becomes very very very large.

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So the way that we would calculate that with calculus is we would say the limit as and approaches infinity

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of our series here so 40 96 to the and over to divided by eight to the two.

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And then this just becomes a limit problem where we have to evaluate this limit.

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So depending on the function you may have to evaluate the limit differently you might just be able to

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use a simple substitution to evaluate the limit you might have to use factoring or conjugate method.

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You might have to use those battels rule if you get an indeterminate form when you plug in this value

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and equals infinity.

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But in this case what we're going to be able to do is actually simplify the fraction itself.

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So what we see here is that we have the exponent and over 2 in the numerator.

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Well that's the same thing as saying four thousand ninety six to one half we take that two out of the

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denominator we put it in the dominator here.

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So essentially factor out the two from the denominator of 40 96 to the one half all race to the end

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power because then when we multiplied x points back together we get an over 2 which is our original

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exponent.

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And of course that's just the same as the square root of 40 96 all raised to the power.

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The square root of 40 96 is 64.

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So we end up with then is 64 to the power in the numerator in the denominator here we can think about

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this as 8 squared and then raised to the power because when we multiply two times and we get two when

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we're back to our original exponent.

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So we just factored out that too.

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Well of course eight squared is 64.

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So we end up with is just 64 to the power.

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So now we're talking about the limit as and approaches infinity of 64 and divided by 64 n.

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Well of course one thing divided by itself is just one.

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So the value of this fraction is 1 so we get the limit as and approaches infinity of 1.

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Well we have no variable to plug in for here.

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So no matter what the value is of n the limit of the function is going to be 1.

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So we can say then that the limit is just equal to 1 because the limit never changes regardless of the

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value of N because there's no variable in this function.

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So the limit is always equal to 1.

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What that means if we were to graph this series so let's put a graph together really quickly up here.

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So if we were to graph this series what it would show us is that the function looks like this.

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If this is why equals 1 so this is one here than the limit or the value of this series is always going

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to be 1 because the function simplifies to 1.

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So essentially our series here is one.

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Therefore the limit of the series is 1.

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Now what happens when we want to find the sum of the series.

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Well remember our series simplified to 1.

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So we would have the sum from unequals 1 to infinity of 1 because 40 96 to the end over 2.

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Divided by eight to the two N is equal to 1.

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So we'd have the same here.

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So then if we plugged unequals 1 into our function we start trying to find the sum here we start expanding

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the sum.

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We plug in any one where there's no variable to plug in for.

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We so just get one if we plug in N equals 2.

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We still just get once we do I wonder that if we plug n equals 3 into our series we still just get one.

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So we'd add 1 to that and we would keep going.

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We would keep adding one on into infinity.

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And so if we keep adding 1 to the sum then obviously this sum is going to be infinite and the sum of

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the series is infinity.

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Another way you could think about that.

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If you just plugged unequals one two three four into this original series here or even into 64 to the

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end over 64 to the end.

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Obviously we would have 64 to the first over 64 to the first plus 64 to the second.

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Over 64 to the second plus 64 to the third over 64 to the third.

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And we could go on like that forever this would be the term that corresponds to an equals 1 and equals

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to an end equals three.

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But obviously here we would just get one plus one plus one.

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And so we would have a sum of infinity.

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So you can see here that the limit of the series is one we can set a limit of the series.

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Here is one.

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But the sum of the series is infinity.

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And so you can see that we get different values for the limit and the sum of the series needed to make

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sure you know how to find both.