Worksheet 7, Math 10560

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1 Worksheet 7, Math 0560 You must show all of your work to receive credit!. Determine whether the following series and sequences converge or diverge, and evaluate if they converge. If they diverge, you must explain why to get credit. X 5 n + n (a) n X 5 n Solution: Because is a geometric series with ration 5 n X 5 n + n series diverges. Therefore, diverges. n >, this geometric (b) X ln( n + n + ) Solution: This is a typical telescoping series. More precisely, let s first consider a partial sum S N = NX ln( n + N n + )= X ln(n +) ln(n +) =ln ln {z } +ln ln 4 {z } =ln ln(n +). +ln4 {z ln 5 } ln(n +) ln(n +) {z } n= n=n Thus, X ln( n + )= lim n + S N =ln ln(n +)=. N! (c) a n =sin( n ) Solution: Looking at the graph of sin(x), we notice that for any even number a and for a < < (a +), thensin( ) > 0, and for an odd number b and b < < (b +) then sin( ) < 0. Now looking at the sequence {n/ } you can check that you can always find integers m and m such that a < m / < (a +) and b < m / < (b+) for a even and b odd. Hence the sequence sin(n/ ) is

2 constantly changing from positive numbers to negative numbers and vice-versa, which implies that if it converges, it has to converge to 0. However, sin(x) =0 only when x is an integer multiple of. Since {n/ } is never approaching multiples of, thenthesequencesin(n/ ) diverges. (d) a n =sin( n ) Solution: Because sin x is a continuous function for all x is a real number, we have lim n! sin( n )=sin(lim n! n )=sin0=0.

3 . (a) Show that f(x) = p x is a continuous, positive, decreasing function on [, ). Solution: f is continuous because it is a composition of continuous functions and the only points at which it is not defined are and -, which are not in our interval. f is clearly positive. f 0 x (x) = < 0on[, ), so f is decreasing on our interval. (x ) (b) Compute R p x dx. Solution: Observe that R p dx R t x t! p x dx. If you re good with hyperbolic trigonometry, you ll recognize straight away that this is the antiderivative of the inverse hyperbolic cosine. The rest of us, however, will have to use a trigonometric substitution. Let x =secu. Then dx =secutan udu, when x =,u =, and when x = t, u =arcsect. Z t lim p t! x Z arcsec t dx t! Z arcsec t t! Z arcsec t t! sec u tan u p sec u du sec u tan u p tan u du sec udu t! ln sec u +tanu t! t! arcsec t! ln sec (arcsec t)+tan(arcsect) ln +sin(arcsect) ln cos (arcsec t) t! ln + p t t t ln + p + p! ln t + p t ln + p t! =ln lim t + p t ln + p t! = ln + p = ln sec +tan Therefore, R p dx = x (c) Determine whether the series P p n is convergent or divergent.

4 Solution: By our work in parts (a) (checking the requirement) and (b) (taking the integral), the series diverges by the Integral Test.

5 . For each of the following series, determine whether they converge, and state why. (a) P sin(n) Solution: The sequence sin(n) does not converge, so the series is divergent by the Divergence Test. (b) P n (c) P Solution: This series converges by the p-test, with p =>. n cos (n) Solution: We can use the Comparison Test. 0 < cos (n) apple (whyiscos (n) never 0?), so n cos (n) apple n, and. P n cos (n) n diverges, and so the given n series diverges also. (d) P 000( 5 4 )n Solution: This is a geometric series with ratio 5 4. But 5 4 > sotheseriesis divergent. You can also see this by the Divergence Test.

6 4. The meaning of the decimal representation of a number 0.d d d... (where the digit d i is on of the numbers 0,,..., 9) is that Show that this series always converges. 0.d d d d 4... = d 0 + d 0 + d 0 + d Solution: We note that d 0 + d 0 + d 0 + d = X d n 4 0 n However, we know that each 0 apple d n apple 0, so 0 apple dn that Thus, by the Comparison Test, 0.d d d d 4... always con- so the series converges. verges. X 0 = n 0 = 0 9 apple 0 = 0 n 0 n 0 n. We now see Another way to see this is to consider the sequence of partial sums, s N = P N d n. 0 n This sequence is clearly increasing, since we are always adding on positive terms, and it is bounded above since 0.d d d d 4... apple, so it must converge, because it is a bounded monotone sequence. The limit is the sum of the series. 5. Use integration to calculate upper and lower bounds for the series P. (Hint: Draw n the graph of the function, and calculate two di erent Riemann sums.) Solution: On the interval (0, ), f(x) = is a continuous decreasing function. x Therefore R we could calculate upper and lower bounds for the series by the integrals f(x) dx and R f(x) dx. These evaluate as lim t! = and lim t t! t = respectively. 8 8

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