(a) Use the sum of the first 10 terms to estimate the sum of...

(a) Use the sum of the first 10 terms to estimate the sum of the series $\Sigma_{n-1}^{\infty} 1 / n^{2} .$ How good is this estimate? (b) Improve this estimate using ( 3) with $n=10$. (c) Compare your estimate in part (b) with the exact value given in Exercise 34 . (d) Find a value of $n$ that will ensure that the error in the approximation $s \approx s_{n}$ is less than $0.001 .$

Transcript

00:01 In this question first of all we will consider a series which is n is equal to 1 to infinity for 1 upon n squared.

00:12 So now according to the question for solution of first part we will find our value for s n which is equal to summation of all values of n is equal to 1 to 10 1 upon n squared that will be equal to 1 by 1 whole square plus 1 upon 2 whole square 1 upon 3 whole square 1 by 4 whole square 1 by 5 whole square then 1 upon 6 whole square and so on till 1 by 10 whole square we have all the terms over here and the values will be as this will be 1 then 1 divided by 4 is 0 .25 added by next term will be as 0 .111 1 1 1 1 1 1 1 1 1 1 and so on.

01:07 The last term will be as 0 .0 and their sum will be as 1 .5 -497677 which can be considered after getting rounding of 1 .9 49768 so here is the sum of first 10 and so we have here our first solution answer for a part now let's move on be part of the question so here we will find our value by using our formula s n plus integration of n plus 1 to infinity for f of x d x as for the inequality as less than or equal to s which is less than or equal to s plus integration of n infinity x d x so now let's put down our values which is 1 .549769 added by value of function which is 11 to infinity for 1 upon x squared dx and when we simplify we will get 1 .54968 added by minus 1 by x and our values for the limits which are 11 to infinity so therefore 1 .549768 minus 1 upon infinity added by 1 upon 11 so this term will be 0 and we will get 1 .5 49768 added by 1 upon 11.

03:03 So then we will get here our value which is equal to 1 .6477 for our value as as an added by n plus 1 infinity for f of x t x and let's calculate now s n plus n to infinity for f f f t x and let's calculate now s n plus n to infinity for function of x d x so with the same way we can calculate this also 1 .5497 8 added by n which is 10 to infinity for 1 upon x squared d x and now it is 1 .549768 plus minus 1 upon x and the limits are 10 to infinity and this will be as again 1 .549768 minus 1 upon infinity plus 1 upon 10 and now this middle term is again 0 so we will get here 1 .649769 so answer for b part will be s will lie in between 1 .64067 let me write it again s will lie in between 1 .64067 and 1 .649769...

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