Calculus/Sequences and Series/Exercises

1. Assume that the n^th partial sum of a series is given by $s_n=2-\frac{1}{3^n}$.

   a) Does the series converge? If so, to what value?

   b) What is the formula for the n^th term of the series?

2. Find the value to which each the following series converges:

   a) $\sum_{n=0}^{\infty} \frac{3}{4^n}$

   b) $\sum_{n=1}^{\infty} \left(\frac{2}{e}\right)^n$

   c) $\sum_{n=2}^{\infty} \frac{1}{n^2-n}$

   d) $\sum_{n=1}^{\infty} \frac{(-1)^n 2^{n-1}}{3^n}$

3. Determine whether each the following series converges or diverges:

   a) $\sum_{n=1}^{\infty} \frac{1}{n^2}$

   b) $\sum_{n=0}^{\infty} \frac{1}{2^n}$

   c) $\sum_{n=1}^{\infty} \frac{n}{n^2+1}$

   d) $\sum_{n=2}^{\infty} \frac{1}{\ln n}$

   e) $\sum_{n=0}^{\infty} \frac{n!}{2^n}$

   f) $\sum_{n=1}^{\infty} \frac{\cos\pi n}{n}$

   g) $\sum_{n=2}^{\infty} \frac{(-1)^n}{n\ln n-1}$

4. Determine whether each the following series converges conditionally, converges absolutely, or diverges:

   a) $\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}$

   b) $\sum_{n=2}^{\infty} \frac{(-1)^n \ln n}{n}$

   c) $\sum_{n=2}^{\infty} \frac{(-1)^n n}{(\ln n)^2}$

   d) $\sum_{n=1}^{\infty} \frac{(-1)^n 2^n}{e^n-1}$

   e) $\sum_{n=1}^{\infty} \frac{(-1)^n}{\sin n}$

   f) $\sum_{n=1}^{\infty} \frac{(-1)^n n!}{(2n)!}$

   g) $\sum_{n=1}^{\infty} \frac{(-1)^n e^{1/n}}{\arctan n}$

Hints

1.  

   a) take a limit

   b) s_n = s_{n − 1} + a_n

2.  

   a) sum of an infinite geometric series

   b) sum of an infinite geometric series

   c) telescoping series

   d) rewrite so that all exponents are n

3.  

   a) p-series

   b) geometric series

   c) limit comparison test

   d) direct comparison test

   e) divergence test

   f) alternating series test

   g) alternating series test

4.  

   a) alternating series test; direct comparison test or integral test

   b) alternating series test; integral test or direct comparison test

   c) divergence test

   d) alternating series test (optional); limit comparison test with geometric series

   e) divergence test

   f) ratio test

   g) divergence test

Answers only

1.  

   a) The series converges to 2.

   b) $a_n=\frac{2}{3^n}$

2.  

   a) 4

   b) $\frac{2}{e-2}$

   c) 1

   d) −1/5

3.  

   a) converges

   b) converges

   c) diverges

   d) diverges

   e) diverges

   f) converges

   g) diverges

4.  

   a) converges conditionally

   b) converges conditionally

   c) diverges

   d) converges absolutely

   e) diverges

   f) converges absolutely

   g) diverges

Full solutions

1.  

   a) The series converges to 2 since:

   $s=\lim_{n\to\infty} s_n=\lim_{n\to\infty} \left(2-\frac{1}{3^n}\right)=2$

   b) $a_n=s_n-s_{n-1}=\left(2-\frac{1}{3^n}\right)-\left(2-\frac{1}{3^{n-1}}\right)=\frac{1}{3^{n-1}}-\frac{1}{3^n}=\frac{3}{3^n}-\frac{1}{3^n}=\frac{2}{3^n}$

2.  

   a) The series is

   $\sum_{n=0}^{\infty} 3\left(\frac{1}{4}\right)^n$

   and so is geometric with first term a = 3 and common ratio r = 1/4. So

   $s=\frac{a}{1-r}=\frac{3}{1-1/4}=4.$

   b) $s=\frac{2/e}{1-2/e}=\frac{2}{e-2}$

   c) Note that

   $\sum_{n=2}^{\infty} \frac{1}{n^2-n} = \sum_{n=2}^{\infty} \frac{1}{n(n-1)} = \sum_{n=2}^{\infty} \left(\frac{1}{n-1}-\frac{1}{n}\right)$

   by partial fractions. So

   $s = \lim_{N\to\infty} s_N = \lim_{N\to\infty} \left(1-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{3}-\frac{1}{4}\right) + \ldots + \left(\frac{1}{N-1}-\frac{1}{N}\right).$

   All but the first and last terms cancel out, so

   $s = \lim_{N\to\infty} \left(1-\frac{1}{N}\right) = 1.$

   d) The series simplifies to

   $\sum_{n=1}^{\infty} \frac{(-1)^n 2^n}{3^n \cdot 2} = \sum_{n=1}^{\infty} \frac{1}{2} \left(\frac{-2}{3}\right)^n,$

   and so is geometric with common ratio r = − 2 / 3 and first term − 1 / 3. Thus

   $s=\frac{-1/3}{1-(-2/3)}=-1/5.$

3.  

   a) This is a p-series with p = 2. Since p > 1, the series converges.

   b) This is a geometric series with common ratio r = 1/2, and so converges since | r | < 1.

   c) This series can be compared to a p-series:

   $\sum_{n=1}^{\infty} \frac{n}{n^2+1} \sim \sum_{n=1}^{\infty} \frac{n}{n^2} = \sum_{n=1}^{\infty} \frac{1}{n}$

   The ∼ symbol means the two series are "asymptotically equivalent"—that is, they either both converge or both diverge because their terms behave so similarly when summed as n gets very large. This can be shown by the limit comparison test:

   $\lim_{n\to\infty} \left( \frac{n}{n^2+1} \div \frac{1}{n} \right) = \lim_{n\to\infty} \left( \frac{n}{n^2+1} \cdot \frac{n}{1} \right) = \lim_{n\to\infty} \frac{n^2}{n^2+1} = 1$

   Since the limit is positive and finite, the two series either both converge or both diverge. The simpler series diverges because it is a p-series with p = 1 (harmonic series), and so the original series diverges by the limit comparison test.

   d) This series can be compared to a smaller p-series:

   $\sum_{n=2}^{\infty} \frac{1}{\ln n} \ge \sum_{n=2}^{\infty} \frac{1}{n}$

   The p-series diverges since p = 1 (harmonic series), so the larger series diverges by the appropriate direct comparison test.

   e) The terms of this series do not have a limit of zero. Note that when n > 1,

   $\frac{n!}{2^n} = \frac{n}{2}\cdot\left[\frac{n-1}{2}\cdot\frac{n-2}{2}\dots\frac{2}{2}\right]\cdot\frac{1}{2} \ge \frac{n}{2}\cdot(1)\cdot\frac{1}{2} = \frac{n}{4}$

   To see why the inequality holds, consider that when n = 2 none of the fractions in the square brackets above are actually there; when n = 3 only 2/2 (which is the same as [n − 1] / 2) is in the brackets; when n = 4 only 3/2 (equal to [n − 1] / 2) and 2/2 (equal to [n − 2] / 2) are there; when n = 5, only 4/2, 3/2, and 2/2 are there; and so forth. Clearly none of these fractions are less than 1 and they never will be, no matter what n > 1 is used. Thus

   $\lim_{n\to\infty} \frac{n!}{2^n} = \infty$

   Therefore the series diverges by the divergence test.

   f) This is an alternating series:

   $\sum_{n=1}^{\infty} \frac{\cos\pi n}{n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n}$

   Since the sequence

   $|a_n|=\frac{1}{n}$

   decreases to 0, the series converges by the alternating series test.

   g) Since the terms alternate, consider the sequence

   $|a_n|=\frac{1}{n\ln n-1}$

   This sequence is clearly decreasing (since both n and ln n are increasing — one may also show that the derivative [with respect to n] of the expression is negative for $n\ge2$) and has limit zero (the denominator goes to infinity), so the series converges by the alternating series test.

4.  

   a) This series alternates, so consider the sequence

   $|a_n|=\frac{1}{\sqrt{n}}$

   Since this sequence is clearly decreasing to zero, the original series is convergent by the alternating series test. Now, consider the series formed by taking the absolute value of the terms of the original series:

   $\sum |a_n|=\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$

   This new series can be compared to a p-series:

   $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \ge \sum_{n=1}^{\infty} \frac{1}{n}$

   Since the smaller series diverges, the larger one diverges. But this means the original (alternating) series was not absolutely convergent. Thus, the original series is only conditionally convergent.

   b) solution to come

   c) solution to come

   d) solution to come

   e) solution to come

   f) solution to come

   g) solution to come