Sequences and Series: practice solutions, College – download the PDF

1 Limit of a rational sequence ★★★

Divide the numerator and denominator by \(n\): \(\dfrac{5-\frac3n}{2+\frac7n}\). Since \(\frac3n\to0\) and \(\frac7n\to 0\), the limit is \(\dfrac{5}{2}\).

2 Sum of a geometric series ★★★

The first term is \(a=3\) and each term is half the previous one, so \(r=\tfrac12\). Since \(|r| < 1\), the sum is \(\dfrac{3}{1-\frac12}=6\).

3 Use the divergence test ★★★

\(\dfrac{n}{n+4}=\dfrac{1}{1+\frac4n}\to 1\neq 0\). By the divergence test, the series diverges.

4 Classify three p-series ★★★

These are p-series. (a) \(p=3 > 1\): converges. (b) \(\dfrac1{\sqrt n}=n^{-1/2}\), so \(p=\tfrac12\le 1\): diverges. (c) \(p=\tfrac23\le 1\): diverges.

5 A repeating decimal ★★★

\(0.272727\ldots = 0.27+0.0027+0.000027+\cdots\), with \(a=0.27\) and \(r=0.01\). The sum is \(\dfrac{0.27}{1-0.01}=\dfrac{0.27}{0.99}=\dfrac{27}{99}=\dfrac{3}{11}\).

6 A telescoping partial sum ★★★

The terms telescope: \(s_N=\left(1-\tfrac12\right)+\left(\tfrac12-\tfrac13\right)+\cdots+\left(\tfrac1N-\tfrac1{N+1}\right)=1-\dfrac{1}{N+1}\). So \(s_4=1-\tfrac15=\tfrac45\). As \(N\to\infty\), \(s_N\to 1\), so the sum is 1.

7 Radius of a simple power series ★★★

This is geometric with ratio \(r=\tfrac{x}{2}\). It converges when \(\left|\tfrac x2\right| < 1\), that is \(|x| < 2\), so \(R=2\). At \(x=1\) the sum is \(\dfrac{1}{1-\frac12}=2\).

8 A bouncing ball ★★★

The first fall is 10 ft. The rebound heights are \(6, 3.6, 2.16,\dots\), a geometric sequence with \(a=6\) and \(r=0.6\); each rebound is traveled twice (up and down).

Sum of rebound heights: \(\dfrac{6}{1-0.6}=15\) ft. Total: \(10+2(15)=40\) ft, which is about 12.2 m.

9 Ratio test with powers ★★★

\(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{(n+1)^2}{3^{n+1}}\cdot\dfrac{3^n}{n^2}=\dfrac13\left(\dfrac{n+1}{n}\right)^2\to\dfrac13 < 1\). The series converges absolutely.

10 Two comparisons ★★★

(a) For \(n\ge 1\), \(\dfrac{1}{n^2+5} < \dfrac1{n^2}\), and \(\sum\frac1{n^2}\) converges, so the series converges.

(b) Compare with \(b_n=\dfrac1n\): \(\dfrac{a_n}{b_n}=\dfrac{n}{2n-1}\to\dfrac12\), a finite positive number. Since \(\sum \frac1n\) diverges, so does \(\sum\frac{1}{2n-1}\).

11 Integral test with an exponential ★★★

Let \(f(x)=xe^{-x^2}\). It is positive and continuous. Its derivative is \(f'(x)=(1-2x^2)e^{-x^2} < 0\) for \(x\ge 1\), so \(f\) is decreasing.

\(\displaystyle\int_1^{\infty} xe^{-x^2}\,dx=\left[-\tfrac12 e^{-x^2}\right]_1^{\infty}=\tfrac{1}{2e}\approx 0.184\), which is finite. So the series converges.

12 How many terms are enough? ★★★

The terms \(\frac1{n^2}\) decrease to 0, so the error after \(N\) terms is at most \(b_{N+1}=\dfrac{1}{(N+1)^2}\).

For \(N=2\): \(\frac19\approx 0.111\), too big. For \(N=3\): \(\frac1{16}=0.0625 < 0.1\). So three terms suffice: \(s_3=1-\tfrac14+\tfrac19=\tfrac{31}{36}\approx 0.861\).

13 Absolute or conditional? ★★★

The terms \(\frac1{\sqrt n}\) decrease to 0, so the alternating series test shows convergence. The absolute series is \(\sum n^{-1/2}\), a p-series with \(p=\tfrac12\le1\), which diverges. So the series is conditionally convergent.

14 A Maclaurin approximation ★★★

Replace \(x\) by \(2x\) in \(e^x\): \(e^{2x}=1+2x+\dfrac{(2x)^2}{2}+\dfrac{(2x)^3}{6}+\cdots=1+2x+2x^2+\tfrac43x^3+\cdots\).

At \(x=0.1\): \(1+0.2+0.02+0.00133=1.22133\). The exact value is \(e^{0.2}\approx 1.22140\), so the error is less than 0.0001.

15 True or false? ★★★

(a) False: \(a_n=\frac1n\to0\) but the harmonic series diverges. (b) True: this is the divergence test. (c) True: absolute convergence implies convergence. (d) False: the terms \(\pm1\) do not tend to 0, and the partial sums keep switching between two values (\(-1\) and \(0\) if the sum starts at \(n=1\)).

16 Interval of convergence ★★★

Ratio test: \(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{|x|}{3}\cdot\dfrac{n}{n+1}\to\dfrac{|x|}{3}\). The series converges for \(|x| < 3\), so \(R=3\).

At \(x=3\): \(\sum\frac1n\), the harmonic series, diverges. At \(x=-3\): \(\sum\frac{(-1)^n}{n}\), which converges by the alternating series test. The interval is \([-3,\,3)\).

17 A telescoping series with partial fractions ★★★

Partial fractions: \(\dfrac{2}{n(n+2)}=\dfrac1n-\dfrac1{n+2}\). The partial sum telescopes, leaving \(s_N=1+\tfrac12-\tfrac1{N+1}-\tfrac1{N+2}\).

As \(N\to\infty\), \(s_N\to\tfrac32\). The sum is \(\tfrac32\).

18 When the ratio test fails ★★★

For \(\frac1n\): \(\dfrac{n}{n+1}\to1\). For \(\frac1{n^2}\): \(\dfrac{n^2}{(n+1)^2}\to 1\). Both give \(L=1\).

But \(\sum\frac1n\) diverges and \(\sum\frac1{n^2}\) converges (p-series). So when \(L=1\) the series can do either: the ratio test is inconclusive.

19 Root test ★★★

The root test gives \(\sqrt[n]{|a_n|}=\dfrac{2n+1}{3n+4}\to\dfrac23 < 1\). The series converges absolutely.

20 Estimating a logarithm ★★★

With \(x=0.1\): \(0.1-0.005+0.000333=0.095333\).

The series alternates with decreasing terms, so the error is at most the next term, \(\dfrac{0.1^4}{4}=0.000025\). The true value \(0.095310\) indeed differs by about \(0.000023\).

21 Differentiating a power series ★★★

Differentiate term by term: \(\dfrac{1}{(1-x)^2}=\sum_{n=1}^{\infty} n x^{n-1}\). Multiply by \(x\): \(\sum n x^n=\dfrac{x}{(1-x)^2}\).

At \(x=\tfrac12\): \(\dfrac{1/2}{(1/2)^2}=2\). So \(\sum\dfrac{n}{2^n}=2\).

22 Factorial and power ★★★

\(\dfrac{a_{n+1}}{a_n}=\dfrac{(n+1)!}{(n+1)^{n+1}}\cdot\dfrac{n^n}{n!}=\dfrac{n^n}{(n+1)^n}=\left(1+\dfrac1n\right)^{-n}\to\dfrac1e\approx0.368 < 1\). The series converges.