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# Consider the series $\sum_{n = 1}^{\infty} n/(n + 1)!.$(a) Find the partial sums $s_1, s_2, s_3,$ and $s_4.$ Do you recognize the denominators? Use the pattern to guess a formula for $s_n.$(b) Use mathematical induction to prove your guess.(c) Show that the given infinite series is convergent, and find its sum.

## a. $\frac{(n+1) !-1}{(n+1) !}$b. $\frac{(k+2) !-1}{(k+2) !}$c. $\sum_{n=1}^{\infty} \frac{n}{(n+1) !}=1$

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##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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Let's find the partial sums for the Siri's for part a s. One more. That's the first term of the Siri's here. So go ahead and plug in and equals one up there we have one half. That's our first term as to all that's just as one plus a two. So here we have one half and then we do A to supply gin and equals two friend. This is AM recall. So we have to over three factorial and then go ahead and simplify this to get five over six. Do we have enough to get the pattern yet? Probably not yet. Let's go and find a few more terms. History. It's just as two plus a three So five over six plus three over for Factorial that'LL take a little work to simplify. But there we should get twenty three or twenty four. One thing I know this is that the numerator is always one less than the denominator, as for is as three plus a for so that's s three, which is twenty three over twenty four plus a four which is for over five factorial. And then here this has to be simplified. So combined some fractions here get a common denominator. We have one nineteen over one twenty. And once again the numerator is one lesson the denominator. So we have to guess a formula for this s end. And whatever the denominator is, the numerator is just one less. I'll just put the same exact thing, but then followed by a minus one. So now I have to describe what's in the denominator. Well, when I plug in one I get to When I played in two I get six When I plug in three, there's a jump all the way to twenty four and then someone twenty these numbers look like they're coming from factorial Sze Here's to Factorial three factorial four factorial, five factorial. So in general is just and plus one factorial and plus one on top is well and then that's a factorial there. So to verify that this formula is correct. Well, go ahead and use party here. This is the mathematical induction. So it's going to the next page and actually verify that this formula is correct. And before we do that, just recall that our our value for us one wass one half So let's go to the next page. So we're claiming that sn is n plus one factorial minus one over and plus one factorial. So we're using mathematical induction. You First. We have to check the base case when we plug in and equals one as one. Let's plug into this formula here, you get to minus one over to factorial, which is one half, and that agrees with the value of S one from the previous page. So that's Kurt, that check. So then we have the inductive step where we go ahead into suppose that this equation is true for some value. Let's call it kay. Some using K It's OK is just some particular value of N for which this equation is true. And now I'd like to show that this is also we have a symbol of formula for escape plus one. So on one hand, as K plus one is escape plus a plus one, you take the previous partial some and then you just add the newest term. And now I could I could describe S k using this formula from our inductive hypothesis. So this is K plus one factorial minus one kaye plus one factorial and then minus and then a K plus one. While I already have a formula for a can. The previous page on the previous page we know a n is and over and plus one fancy Uriel. So use that formula with n equals K plus one. Now we have this new expression here Would want to simplify this eventually. What we want is this. We want this formula to still be true when we replace came with K plus one. So after we replace it on the left than on the right, a inside this is what we have. So that's what we want to have for escape plus one. So we started with escape plus one over here on the left. We're using the definition of the partial sons Will simple find this and eventually we want to end with this expression over here in the blue. So we're almost there. We have a few steps left. Let's just combined these fractions. Yeah, Kaye plus one factorial minus one. And then there's K plus two and then plus que plus one all over Cape Lis. Two factorial. So I did here was multiplying top and bottom of the red by K plus two And then we can go out and simplify this. So if I multiply that K plus two on the inside that becomes K plus two factorial and then I have minus okay plus two and then plus que plus one. So that's just distributing the K plus two to the negative one as well, all over K plus two factorial and then just simplified that numerator minus and then just case cancel. We have minus two plus one. So that's minus one all over K plus two factorial. And yes, this is exactly what we wanted. This means that the formula is still true when we replace Kay from our hypothesis with K plus one. So that verifies party that our guests, that our formula for SN was correct. Now let's go ahead to do party. So for part c, we have that that it's conversion. They're telling us to suppose that this thing converges So let's write this as a limit. A cz n goes to infinity. Let's not use and let's use a different letter there. Let's save Kei goes to infinity of S k. So we're taking the limit is the partial sums. However, from part B, we just found the formula for S K. That was K plus one factorial minus one over Kaye plus one factorial. So this is just from using part B to go from here to here. Then this limit can be evaluated. But it might help to rewrite this by Let's go ahead and divide. It's happened bottom by the drama, the denominator. So when we do this, or, in other words, just split this into two fractions. So we have one fraction when they can't allow minus one over Kaye, plus one factorial. And then as we take the limit, this denominator gets really, really large in the limit. So the fraction goes to zero because there was only a one up there, and then we just have one, and that's the sum of the series, and this is our final answer.

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##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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