Series 215 problem 04.mws

1 . Comparison tests.

Given the series Sum(1/(n^4+n^2),n = 1 .. infinity) . We may argue that this series is convergent by using the comparison tests. First, we note the series under consideration is a positive term series, and 1/(n^4+n^2) <= 1/(n^4) for all 1 <= n . Recall Sum(1/(n^4),n = 1 .. infinity) converges since it is a p-series with p = 4 . Therefore by the Comparison Test, the series Sum(1/(n^4+n^2),n = 1 .. infinity) is convergent. Alternatively we could argue from the Limit Comparison Test: Notice that limit((n^4+n^2)/(n^4),n = infinity) = 1 , and therefore the series Sum(1/(n^4+n^2),n = 1 .. infinity) is convergent. Now, since we know that the series converges, we will use Maple to sum the first 10 terms of this series, this is easy:

> sum(1/(n^4+n^2),n=1..10);

122131997547449/215030621332800

Perhaps an approximation is ok in this case.

> evalf(sum(1/(n^4+n^2),n=1..10));

.5679749088

Now, how close is this number to the actual sum. The error in this case is underneath int(1/(x^4),x = 10 .. infinity) = 1/(3*10^3) .

> evalf(sum(1/(n^4+n^2),n=1..infinity));

.5682600205+0.*I

And we see that our above estimate above is within .003 of the actual sum.

Submission:

Use the sum of the first 10 terms to approximate the sum of the given series. Use the Integral test remainder estimate to get an upper bound on the error. Be careful to justify that the series converges before approximating the sum.

(a) Sum((1+cos(n))/(n^5),n = 1 .. infinity)

(b) Sum(n/((n+1)*3^n),n = 1 .. infinity)

Submission worksheet:

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2. Alternating series test.

The series sum((-1)^(n-1)/(n^p),n = 1 .. infinity) is an alternating series. We want to determine the value of p so that this series is convergent. The alternating series test is helpful here. First notice that 1/((n+1)^p) <= 1/(n^p) for all n where 1 <= n as long as 0 <= p . This means the sequence of the absolute values of the terms of the series is decreasing as long as 0 <= p . Secondly, limit(1/(n^p),n = infinity) = 0 if 0 < p . Therefore, the conditions of the Alternating Series Test are met as long as 0 < p , thus the series sum((-1)^(n-1)/(n^p),n = 1 .. infinity) is convergent. Now let us be more specific and choose p = 2 and estimate the sum in this case with an error no bigger than 0.01. By the Alternating Series Estimation Theorem, we know that this surely happens if 1/((n+1)^2) <= .1e-1 . This is a fairly easy inequality to solve, and we get 9 <= n . So, summing the first 9 terms of the series will give an error no greater than 0.01.

> evalf(sum((-1)^(n-1)/n^2,n=1..9));

.8279621756

Submission:

(a) Approximate the sum of the series Sum((-1)^(n+1)/(n^4),n = 1 .. infinity) within an error of 0.001

(b) Calculate the first 10 partial sums of the series Sum((-1)^(n-1)/(n^3),n = 1 .. infinity) and graph both the sequence of terms and

the sequence of partial sums on the same set of axes. Estimate the error in using the 10th partial sum to

approximate the total sum.

Argue carefully that the series above are convergent before estimating their sums.

Submission worksheet:

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