Math 132: Calculus for Physical and Life Sciences 2 Problem Set 8 Due Friday, April 11, 2008, at the beginning of class. Solutions

1. Determine whether each of the following sequences converges or diverges. If it converges, find its limit.

(a) a_{n} = \frac{1 − n^{2}}{2 + 3n^{2}}

Solution: lim_{n→∞} \frac{1 − n^{2}}{2 + 3n^{2}} = lim_{n→∞} \frac{\frac{1}{n^{2}} − 1}{\frac{2}{n^{2}} + 3} = −\frac{1}{3}.

Or use l’Hospital’s rule twice: lim_{n→∞} \frac{1 − n^{2}}{2 + 3n^{2}} = lim_{x→∞} \frac{1 − x^{2}}{2 + 3x^{2}} =^{l′H} lim_{x→∞} \frac{−2x}{6x} =^{l′H} lim_{x→∞} \frac{−2}{6} = −\frac{1}{3}

(b) a_{n} = 1 + (−1)^{n}

Solution: a_{n} = 0 if n is odd and a_{n} = 2 if n is even. The sequence does not approach a single number and thus it diverges.

(c) a_{n} = \frac{sin n}{3^{n}}

Solution: Use the Squeeze Theorem. −1 ≤ sin n ≤ 1 for all n. Thus −\frac{1}{3^{n}} ≤ \frac{sin n}{3^{n}} ≤ \frac{1}{3^{n}}.

Since lim_{n→∞} −\frac{1}{3^{n}} = lim_{n→∞} \frac{1}{3^{n}} = 0, we have that lim_{n→∞} \frac{sin n}{3^{n}} = 0.

(d) a_{n} = \frac{(ln n)^{2}}{n}

Solution: lim_{n→∞} \frac{(ln n)^{2}}{n} = lim_{x→∞} \frac{(ln x)^{2}}{x} =^{l′H} lim_{x→∞} \frac{2(ln x)\frac{1}{x}}{1} = lim_{x→∞} \frac{2 ln x}{x} =^{l′H} lim_{x→∞} \frac{2}{x} = 0.

(e) a_{n} = \frac{2^{n} + 1}{e^{n}}

Solution: lim_{n→∞} \frac{2^{n} + 1}{e^{n}} = lim_{n→∞} (\frac{2}{e})^{n} + lim_{n→∞} (\frac{1}{e})^{n} = 0 since both \frac{2}{e} and \frac{1}{e} are between 0 and

1.

(f) a_{n} = ln(2n + 1) − ln(2n − 1)

Solution: lim_{n→∞} a_{n} = lim_{n→∞} ln (\frac{2n + 1}{2n − 1}) = ln ( lim_{n→∞} \frac{2n + 1}{2n − 1}) = 0.

2. Determine if each infinite series converges or diverges. If it converges, find its sum. (Be careful about the first term of the series.)

(a) 1 + e^{−1} + e^{−2} + e^{−3} + … + e^{−n} + … Solution: ∑_{n=0}^{∞}(e^{−1})^{n} is the geometric series with ratio \frac{1}{e} and first term 1. It is convergent and its sum is \frac{e}{e − 1}.

(b) ∑_{n=0}^{∞} \frac{3^{n} − 2^{n}}{4^{n}}

Solution: ∑_{n=0}^{∞} \frac{3^{n}}{4^{n}} = ∑_{n=0}^{∞} (\frac{3}{4})^{n} = 4 (geometric series with ratio 3/4 and first term 1). ∑_{n=0}^{∞} (\frac{2}{4})^{n} = 2 (geometric series with ratio 1/2 and first term 1).

Thus ∑_{n=0}^{∞} \frac{3^{n} − 2^{n}}{4^{n}} = ∑_{n=0}^{∞} \frac{3^{n}}{4^{n}} − ∑_{n=0}^{∞} \frac{2^{n}}{4^{n}} = 4 − 2 = 2

(c) ∑_{n=0}^{∞} (\frac{100}{99})^{n}

Solution: This is a geometric series with ratio 100/99 > 1. It diverges.

(d) ∑_{n=0}^{∞} (\frac{99}{100})^{n}

Solution: This is a geometric series with ratio 99/100 < 1 and first term 1. It converges to 100.

(e) ∑_{n=1}^{∞} (\frac{π}{e} )^{n}

Solution: This is a geometric series with ratio π/e > 1. It diverges.

(f) ∑_{n=1}^{∞} (\frac{2}{n} − \frac{1}{2^{n}} )

Solution: ∑_{n=0}^{∞} \frac{2}{n} = 2 ∑_{n=0}^{∞} \frac{1}{n} is divergent (twice the harmonic series). ∑_{n=0}^{∞} \frac{1}{2^{n}} converges (geometric series with ratio 1/2). Thus ∑_{n=1}^{∞} (\frac{2}{n} − \frac{1}{2^{n}} ) diverges. ( If ∑_{n=1}^{∞} (\frac{2}{n} − \frac{1}{2^{n}} ) were convergent, ∑_{n=0}^{∞} \frac{2}{n} = ∑_{n=1}^{∞} (\frac{2}{n} − \frac{1}{2^{n}} ) + ∑_{n=0}^{∞} \frac{1}{2^{n}} would be convergent as the sum of two convergent series.)

(g) ∑_{k=1}^{∞} (\frac{k + 1}{2k + 4})

Solution: Since lim_{k→∞} \frac{k + 1}{2k + 4} = \frac{1}{2} ≠ 0, the series diverges by the Divergence Test.

(h) ∑_{n=1}^{∞} ln (\frac{n + 1}{n})

(Hint: Consider several partial sums and use properties of logarithms to simplify them.)

Solution: Use the fact that ln (\frac{n + 1}{n}) = ln(n + 1) − ln n. Consider the n-th partial sum

s_{n} = (ln 2 − ln 1) + (ln 3 − ln 2) + (ln 4 − ln 3) + ⋯ + (ln n − ln(n − 1)) + (ln(n + 1) − ln n). All terms cancel except for ln 1 and ln(n + 1). Thus s_{n} = ln(n + 1) − ln 1 = ln(n + 1). Since lim_{n→∞} s_{n} = ∞, the series diverges.

3. Find the rational number represented by 0.2525252525 . . ..

(Hint: Think of 0.2525252525 … as a geometric series.)

Solution: 0.2525252525 … = \frac{25}{100} + \frac{25}{(100)^{2}} + \frac{25}{(100)^{3}} + ⋯ This is the geometric series with

ratio 1/100 and first term 25/100. It is convergent and its sum is

\frac{\frac{25}{100}}{1 − \frac{1}{100}} = \frac{25}{99}.

Thus 0.2525252525 … = \frac{25}{99}.

4. Suppose the government spends $1 billion and that each recipient spends 90% of the dollars that he or she receives. In turn, the secondary recipients spend 90% of the dollars they receive, and so on. How much total spending results from the original injection of $1 billion into the country?

Solution: The total spending (in Dollars) equals 10^{9} + 10^{9} · 0.9 + 10^{9}(0.9)^{2} + 10^{9}(0.9)^{3} + ⋯ =

\frac{10^{9}}{1 − 0.9} = 9 · 10^{9} since this is a geometric series with ratio 0.9 and first term 10^{9}. Thus the

total spending is $10 billion.

5. Use the integral test to decide whether each of the following series converges or diverges.

(a) ∑_{n=0}^{∞} \frac{n}{n^{2} + 1}

Solution: Let f(x) = \frac{x}{x^{2} + 1}. By the quotient rule (check!) f′(x) = \frac{1 − x^{2}}{(x^{2} + 1)^{2}} . For x > 1 this

is negative and thus f is decreasing. The function f is also continuous and positive for x > 0. We can apply the integral test (use the u-substitution u = x^{2} + 1, du = 2x dx in the integral).

∫_{1}^{∞} \frac{x}{x^{2} + 1} dx = lim_{b→∞} ∫_{1}^{b} \frac{x}{x^{2} + 1} dx = lim_{b→∞} \frac{1}{2} ∫_{2}^{b^{2}+1} \frac{1}{u} du = lim_{b→∞} \frac{1}{2}(ln(b^{2} + 1) − ln 2) = ∞.

Since the integral is divergent, the series is divergent as well.

(b) ∑_{n=0}^{∞} \frac{n}{e^{n^{2}}}

Solution: Let f(x) = \frac{x}{e^{x^{2}}} . We have f′(x) = \frac{e^{x^{2}} − 2x^{2}e^{x^{2}}}{(e^{x^{2}})^{2}} = \frac{e^{x^{2}}(1 − 2x^{2})}{e^{2x^{2}}}. Since f′(x) < 0 for

x ≥ 1, the function f is decreasing. The function f is also continuous and positive for x > 0 and we can apply the integral test (use the u-substitution u = x^{2}, du = 2x dx in the integral).

∫_{1}^{∞} \frac{x}{e^{x^{2}}} dx = lim_{b→∞} ∫_{1}^{b} \frac{x}{e^{x^{2}}} dx = lim_{b→∞} \frac{1}{2} ∫_{1}^{b^{2}} e^{−u} du = lim_{b→∞} \frac{1}{2}(−e^{−b^{2}} + e^{−1}) = \frac{1}{2}e^{−1} = \frac{1}{2e}

Since the integral is convergent, the series converges as well.

(c) ∑_{n=2}^{∞} \frac{1}{n ln n}

Solution: Let f(x) = \frac{1}{x ln x}. Since the denominator is increasing, the function is decreasing. It is positive and continuous for x > 1 and we can apply the integral test (use the u-substitution u = ln x, du = 1/x dx in the integral).

∫_{2}^{∞} \frac{1}{x ln x} dx = lim_{b→∞} ∫_{2}^{b} \frac{1}{x ln x} dx = lim_{b→∞} ∫_{ln 2}^{ln b} \frac{1}{u} du = lim_{b→∞}(ln(ln b) − ln(ln 2)) = ∞.

Since the integral diverges, the series diverges as well.

(d) ∑_{n=1}^{∞} \frac{n}{e^{n}}

Solution: Let f(x) = \frac{x}{e^{x}}. Since f′(x) = \frac{e^{x}(1 − x)}{e^{2x}} is negative for x > 1, the function f is decreasing for x > 1. Also, for x > 0, f is continuous and positive. We can apply the integral test (use integration by parts in the definite integral: u = x du = 1 dx and dv = e^{−x} dx, v = −e^{−x}).

∫_{1}^{∞} \frac{x}{e^{x}} dx = lim_{b→∞} ∫_{1}^{b} xe^{−x} dx = lim_{b→∞}(−xe^{−x}|_{1}^{b} + ∫_{1}^{b} e^{−x} dx = lim_{b→∞}(−be^{−b} − e^{−b} + 2e^{−1}).

By l’Hospital’s rule lim_{b→∞} be^{−b} = lim_{b→∞} \frac{b}{e^{b}} = lim_{b→∞} \frac{1}{e^{b}} = 0. Thus ∫_{1}^{∞} \frac{x}{e^{x}} dx = 2e^{−1}. Since the

integral is convergent, the series is convergent as well.

(e) ∑_{n=1}^{∞} ln (1 + \frac{1}{n^{2}} )

(Hint: integrate by parts.)

Solution: Let f(x) = ln (1 + \frac{1}{x^{2}} ). It is a positive, continuous and decreasing function for x ≥

1 and we can apply the integral test. We will first solve ∫ ln (1 + \frac{1}{x^{2}} ) dx using integration by

parts. Let u = ln (1 + \frac{1}{x^{2}} ) and dv = dx. Then du = \frac{1}{1 + \frac{1}{x^{2}}} · \frac{−2}{x^{3}} dx = −\frac{2}{x(x^{2} + 1)} dx and v =

x. Then ∫ ln (1 + \frac{1}{x^{2}} ) dx = x ln (1 + \frac{1}{x^{2}} )+2 ∫ \frac{1}{1 + x^{2}} dx = x ln (1 + \frac{1}{x^{2}} )+2 arctan x+C.

Then

∫_{1}^{∞} ln (1 + \frac{1}{x^{2}} ) dx = lim_{b→∞} ∫_{1}^{b} ln (1 + \frac{1}{x^{2}} ) dx = lim_{b→∞} (b ln (1 + \frac{1}{b^{2}} ) + 2 arctan b − ln 2 + 2 arctan 2)

We have lim_{b→∞} b ln (1 + \frac{1}{b^{2}} ) = lim_{b→∞} \frac{ln (1 + \frac{1}{b^{2}} )}{\frac{1}{b}} =^{l′H} lim_{b→∞} \frac{_{\frac{1}{1+ \frac{1}{b^{2}}}} · \frac{−2}{b^{3}}}{−\frac{1}{b^{2}}} = lim_{b→∞} \frac{2}{b(1 + \frac{1}{b^{2}} )} = 0 and

lim_{b→∞} arctan b = π/2.

Therefore the integral converges and so does the series.

6. Determine whether the given series converges or diverges. Quote general results to justify your answers.

(a) ∑_{n=1}^{∞} \frac{1}{(\sqrt{n})^{5}}

Solution: ∑_{n=1}^{∞} \frac{1}{(\sqrt{n})^{5}} = ∑_{n=1}^{∞} \frac{1}{n^{5/2}} . This is the p-series with p = 5/2 > 1 and is therefore

convergent.

(b) ∑_{n=1}^{∞} \frac{n + 1}{n^{2}}

Solution: Since ∑_{n=1}^{∞} \frac{1}{n} diverges (harmonic series) and ∑_{n=1}^{∞} \frac{1}{n^{2}} convergent (p-series with p = 2),

the given series diverges.

7. Show that the series ∑_{n=2}^{∞} \frac{1}{n(ln n)^{p}} converges if p > 1 and diverges if p ≤ 1.

Solution: If p = 1 this is the series of problem 5.c) which diverges. Suppose p ≠ 1. Since

f(x) = \frac{1}{x(ln x)^{p}} is positive, continuous and decreasing for x > 1 we use the integral test. In

the integral we use the substitution u = ln x, du = 1/x dx.

∫_{2}^{∞} \frac{1}{x(ln x)^{p}} dx = lim_{b→∞} ∫_{2}^{b} \frac{1}{x(ln x)^{p}} dx = lim_{b→∞} ∫_{ln 2}^{ln b} \frac{1}{u^{p}} du = lim_{b→∞} \frac{u^{−p+1}}{−p + 1}|_{ln 2}^{ln b} =

= lim_{b→∞} (\frac{(ln b)^{−p+1}}{−p + 1} − \frac{(ln 2)^{−p+1}}{−p + 1} )

If p > 1, then lim_{b→∞}(ln b)^{−p+1} = 0 and if p < 1, then lim_{b→∞}(ln b)^{−p+1} = ∞. Therefore, the integral converges is p > 1 and it diverges if p < 1. By the integral test and problem 5.c), the series converges if p > 1 and it diverges if p ≤ 1.

8. For each of the following series explain why the integral test does not apply.

(a) ∑_{n=1}^{∞} e^{−n} sin n

Solution: The integral test does not apply because the function f(x) = e^{−x} sin x is not positive (for example it takes negative values at each odd multiple of π/2) and not decreasing (f′(x) = e^{−x}(cos x − sin x) is not always negative - for example, it is positive at all even multiples of π).

b) ∑_{n=1}^{∞} \frac{2 + sin n}{n^{2}}

Solution: Here the function f(x) = \frac{2 + sin x}{x^{2}} is positive for all x ≠ 0 (because −1 ≤ sin x ≤ 1).

However, the function is not decreasing since f′(x) = \frac{x^{2} cos x − 2x(2 + sin x)}{x^{4}} is not always negative (it is positive for all even multiples of π greater than 4).