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# Euler also found the sum of the $p-$ series with $p = 4:$$\zeta (4) = \displaystyle \sum_{n = 1}^{\infty} \frac {1}{n^4} = \frac {\pi^4}{90}$Use Euler's result to find the sum of the series.(a) $\displaystyle \sum_{n = 1}^{\infty} \left( \frac {3}{n} \right)^4$(b) $\displaystyle \sum_{k = 5}^{\infty} \frac {1}{(k - 2)^4}$

## a) $\frac{9 \pi^{4}}{10}$b) $\frac{\pi^{4}}{90}-\frac{17}{16}$

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##### Top Calculus 2 / BC Educators   ##### Samuel H.

University of Nottingham ##### Michael J.

Idaho State University

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We're given the sum of the P series here, where people's four given my Oiler and this is Piper, the fourth over ninety and would like to use this facto compute the Siri's and part A and B. So for part A. If you just go ahead and simplify the fraction by raising its of the fourth hour now, since the syriza Pierre converges, we can go ahead and pull out the eighty one here. So this is because this is important here. This's by a serum since eighty one is constant and the Siri's that that we mentioned the beginning convergence. So if you know a Siri's conversions, you can go ahead and pull out constant terms like we did here and now. Since we know this sum from the given information no, we just plug that in for the song and then simplify eighty one over ninety. That's nine over ten and then we still have our pipe of the fourth. So that'LL be our Let me go ahead and clean that up a little. That'LL be our answer for party, eh? Try to leave some room here for part B Now for part B. First thing is is we want the denominators to be just one term. So let's go ahead and replace K minus two Once said that equal to and because then that allow us to right the denominator is one over into the fourth. However, we have to change the summation here because case starts at five. Then that means and his three So the sum will start from N equals three and go to infinity. Now this looks close to the original, but the original starts at one. So what I'LL do here is I'LL go ahead and I'm gonna add ins of tracks in terms here. I'm going to go ahead and include the first two terms corresponding to an equals one into and then I'll make up for by subtracting those terms that I added right off. So in other words, this is just one over one to the fourth one over two to the fourth, plus all the terms after two. That's the relationship between the two sons. And that explains how I did this step here. Now finally, this's the given, some from the beginning over up here from Oiler Pint of the four over ninety and then here you have minus one and then minus one over sixteen. So just combined those fractions, giving you minus seventeen over sixteen, and there's our answer for part B.

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##### Top Calculus 2 / BC Educators   ##### Samuel H.

University of Nottingham ##### Michael J.

Idaho State University

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