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JH
Numerade Educator

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Problem 34 Medium Difficulty

Use the sum of the first 10 terms to approximate the sum of the series. Estimate the error.
$ \displaystyle \sum_{n = 1}^{\infty} \frac {e^{1/n}}{n^4} $

Answer

$R_{10} \approx .000359364$

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Video Transcript

let's use the some of the first ten terms to approximate the sum of the Siri's. So on the left over here, this is the sum of the Siri's, and that's approximately equal to us. Ten as ten. We know that's just the sum after using ten terms. So let's go to my next cabin, Wolfram Alpha, where, if you can see, I approximated the S ten the some of the first ten terms, so you could pause the screen. Write down a few decibels here for this approximation. So now that's are approximate. And now we want to estimate the ear as a result of using ten terms. So they're here is that most are ten. That here, this is the remainder. After using ten terms, that's a notation in the book. Now, before we go on in this direction, let's define affects to be one over X. Excellent fourth. So notice that if it's positive here, we're looking at X bigger than or equal to one because of the starting point is one and equals one. So if this positives on his ex is bigger than your equal to one, we could see that F is continuous because the numerator and denominator Robles continuous, and then we can see that f is decreasing. So just show this part. We need to see that the derivative is negative. So here is the question rule. Then swear that denominator. So the denominator is always positive, but we can see that the numerator is always negative. We can pull out the negativity and then we're less with X squared. Plus for X Cube. That's Herman. The parentheses is positive that the nominators positive. However, this negative is always a negative number, So this thing is less than zero. That means efforts also decreasing. So here we would to show convergence for the original Siri's would use the integral test. And so we're using the upper bound for the ear that's given by the inner will test. So here the upper bound would be the integral from ten to Infinity E one over X over X for the fourth and going head and evaluating this in a computer. This is approximately point zero zero zero three five nine and then three, six four so of repeated that point zero zero zero three five nine three six four. So the ear is no larger than this decimal over here. So that's our estimate in our final answer