Solutions to Assignment-11

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1 Solutions to Assignment-. (a) For any sequence { } of positive numbers, show that lim inf + lim inf an lim sup an lim sup +. Show from this, that the root test is stronger than the ratio test. That is, if the ratio test works, the root test will also work. But, as we saw in class, the converse is not true. That is, there are situations in which the root test works, but the ratio test is inconclusive. Solution: We will show that lim sup an lim sup +. The other inequalities also follow in a similar fashion. Denote L = lim sup n a and U = lim sup n+. We proceed by contradiction, so suppose L > U. Let β (U, L). Then there exists an N such that for all n > N, a n+ < β, or equivalently, + < β a N. Inductively, one can conclude that < β n N a N. That is, for any n > N, /n < β N/n a N /n. Now N is fixed, so taking limsup on both sides, since lim β N/n a N /n = β, we see that lim sup /n β, which is a contradiction. (b) Let = n n /n!. Show that lim an = e. Solution: Note that + = (n + )n+ n! (n + )n+ (n + )n = = (n + )!nn (n + )nn n n = ( + ) n e. n a So lim inf n+ a = lim sup n+ = e. But then by the chain of inequalities in the first part the middle two terms are also equal, that is, lim inf n =

2 lim sup n = e, and so lim an = e. 2. The aim of this question is to prove the integral test for series convergence. Let f(x) be a non-negative, decreasing function on [, ) such that lim f(x) = 0. x Let = f(n), and for n =, 2,, denote n s n = a k, I n = k= n f(x) dx. Recall that since the function is non-negative, the improper integral f(x) dx converges, if and only if the sequence I n is bounded. (a) Show that f(n) + I n s n f() + I n. Hint. Divide the interval [, n] into n intervals of unit size, and use the right-end point and left-end point approximations to the integral. Remember that the function is decreasing. Solution: We can write For any x [k, k + ], n I n = k= k+ k f(x) dx. f(k + ) f(x) f(k). Integrating the first inequality from k to k +, and so or equivalently, f(k + ) k+ k f(x) dx, n I n f(k + ) k= = s n f(), s n I n + f(). This proves the right side of the inequality. For the left side, we integrate the second inequality above, and obtain that or equivalently n I n f(k) = s n f(n), k= I n + f(n) s n. 2

3 (b) Show that converges if and only if f(x) dx converges. Solution: Firstly recall that f(x) dx converges if and only R lim R f(x) dx exists. Since f(x) 0, this limit exists if and only if R f(x) dx is bounded by a constant independent of R. Or equivalently, f(x) dx converges {I n } is bounded. Now the conclusion follows from the following chain of equivalences an converges {s n } converges {s n } is bounded (since 0) {I n } is bounded f(x) dx converges. (c) Use this to show that n p converges if and only if p >. Solution: Follows from part(b) and the fact (proven in class) that x p dx converges if and only if p >. (d) What are the values of p, q for which the series converges? n=2 n p (ln n) q Solution: There are many cases we need to deal with. Before that we start with the following simple observation, that for any α > 0, there exists an N > 0 (possibly depending on α) such that ln(n) < n α for all n > N. This follows from the fact that lim ln(n)/n α = 0, which itself can be proved by an application of L Hospital. with this out of the way, we have the following three cases. p >. In this case if q 0, then since ln(n) for all n > 3, n p (ln(n)) q < n p 3

4 for n > 3. So by comparison theorem, the series converges. If on the other hand, q < 0, let r = q. Then r > 0, and (ln n)r n p = (ln n) q n p. By the above observation, for any α > 0, there exists and N such that for any n > N, (ln n) r < n αr, and so (ln n) r n p < n p αr. Since p >, we can choose α > 0 small enough so that p αr >. For such a choice of α, the series n αr p converges. But then by the comparison test, the original series converges. To sum up, in this case, the series converges no matter what the value of q is. p =. Here the series reduces to n(ln n) q. Let f(x) = /x(ln x) q. Then f is on-negative decreasing function on [2, ). By the integral test, the given series converges if and only if 2 f(x) dx converges. But by change of variables u = ln x, we see that R 2 f(x) dx = ln R ln 2 du u q, and so the integral, and hence the series, converges if and only if q >. p <. In this case if q 0, for any n > 3, n p (ln n) q > n p. Since n p diverges for p <, by the comparison test, the given series also diverges. If q > 0, then for any α > 0, there exists an N such that for any n > N, (ln n) q < n αq, and so n p (ln n) q > n p+αq. Since p <, one can choose α > 0 small enough so that p + αq <, and then the series n p+αq diverges. Again by comparison test, the original series diverges. So to sum up, if p <, the series diverges no matter what the value of q. So finally the series converges if and only if (p, q) {p >, q R} {p =, q > }. 4

5 3. Again, let f,, s n and I n be as in question 2 above, and denote d n = s n I n. (a) Show that 0 f(n) d n f(). Solution: This follows directly from the first part of the previous problem. (b) Show that d n converges to a limit. Solution: We calculate d n d n+ = s n s n+ I n + I n+ = f(n + ) + n+ n f(t) dt. Since the function is decreasing, for any t [n, n + ], f(t) f(n + ), so this shows that the term above is positive, thereby implying that {d n } is a decreasing sequence. From part(a) this sequence is also bounded, and hence must converge. (c) If D = lim d n, show that Hint. Show that 0 d n D f(n). d n d n+ f(n) f(n + ). Solution: From the proof for the above part, 0 d m d m+ f(m + ) + m+ m f(t) dt f(m) f(m + ), since f(t) f(m) for all t [m, m + ]. We observe that the sequences on both sides of the inequality are telescoping. So if we add from m = n to infinity, we will obtain since f(x) decreases to zero. (d) Show that 0 d n lim d m f(n) lim f(m) = f(n), m m γ n = n k= k ln n converges to a limit γ (0, ). The constant γ is called the Euler constant. 5

6 Solution: Apply the above two parts to f(x) = /x. This decreases to zero. From the first part 0 γ n, for all n. And hence the limit is between (0, ). (e) Using part (c) with n = 4, show that In actuality, γ < γ < 0.7. Solution: By part(c), γ n n γ γ n. With n = 4, we can compute that γ n In particular 0.69 < γ n < 0.7. So which gives us the required estimate < γ < 0.7 6