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Numerade Educator



Problem 68 Hard Difficulty

If the $ n $th partial sum of a series $ \sum_{n = 1}^{\infty} a_n $ is $ s_n = 3 - n2^{-n}, $ find $ a_n $ and $ \sum_{n = 1}^{\infty} a_n. $


$a_{n}=\left\{\begin{array}{ll}{2.5} & {\text { if } n=1} \\ {\frac{n-2}{2^{n}}} & {\text { if } n \geq 2}\end{array}\right.$
$\sum_{n=1}^{\infty} a_{n}=3$


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Video Transcript

were given that the end partial. Some of this entire Siri's is given by the following formula here. And then we like to find a end as well as the whole sum. So recall by definition, s n is just the sum of the A's. And in this case, we're starting at one here. This index tells us for the star, and we just add all the way up to a M now on one knee, and we'd like to find this and here. So one way too soft for this and is to notice the A N is equal to s and minus s N minus one. Now, the way to see this, we know what s and equals. We know Ascend minus one equals. And actually I could go back to the previous line up here and I could write us and is all the way Teo and minus one. And then I also add one more day and there. So when I go ahead and subtract s and minus s n minus one Well, go ahead and cancel all the a's all the way up into an a minus one, and then we're left over with Anne so that verifies this equation over here. And now, since we have a formula for a reason, we could go ahead and use this. So we have three minus and over to the end. This's for us in. But then we also have minus and then s n minus one. So dis replace all the ends and the formula within minus one. And then this could be simplified. We could cancel the threes and then we have an minus one over two to the n minus one minus and over to the end. And if you really want to come and denominator there, you could go ahead and multiply top and bottom of this left term by two. And then you have two and minus one minus and over to the end, so and minus one over to the end. So finally, all that work and that's the find the formula for a and that's the first part of this question. So a samen is equal to and minus one divided by two the and power Now we'LL go ahead and do this last part is actually less work. Well, possibly less work here, so we can do is right this infinite sum as a limit. So this sunk. We could always pull out the limit. Let's say Que goes to infinity and then we only do the cave partial some here. So in the product disease right here, that's just equal to escape. By definition, by definition, as cases, some of the first K. And now we have a formula for the limit of S K. Well, we have a formula for escape here, So now we need to evaluate the limit of three minus and over to the end. So let's go to the next page for this one. So for something like this, we we technically do have this and determinant form here. This is infinity over infinity. And so, technically, let me use the letter X here, and I'Ll explain why in one woman if you want to use low Patel's rule, then you need differential functions and sequences air not differential because they're not there only defined in natural numbers. So they're not continuous, not differential. So here I'm just replacing and with X. So I gotta have all real numbers. X could be any real number. And then for something like this, we can use lope with house rule because we have infinity over infinity. As we pointed out, the derivative of the exes, just one. And then the derivative of the denominator is to to the ex Ellen of two. And then, as I take the limit here, I get one over Infinity Ellen's who is just a positive number. So this is your zero. So that tells us in the limit the end over, too. And it just goes to zero and that the final answer and the limit we have three minus zero equals three. And that was our formula for the entire sum of the Siri's for one to infinity of AM, and that's your final answer.