Calculus/Infinite series/Exercises

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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?
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) $\sum_{n=1}^{\infty} \frac{(-1)^n 2^{n-1}}{3^n}$
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) $\sum_{n=3}^{\infty} \frac{(-1)^n}{n\ln n-1}$
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) $\sum_{n=1}^{\infty} \frac{(-1)^n e^{1/n}}{\arctan n}$

[edit] Hints

a) take a limit

b) $s n = s n - 1 + a n$
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
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
a) direct comparison test

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

c) divergence test

d) alternating series test; limit comparison test

e) divergence test

f) ratio test

g) divergence test

[edit] Answers only

a) The series converges to 2.

b) $a_n=\frac{2}{3^n}$
a) 4

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

c) 1

d) −1/5
a) converges

b) converges

c) diverges

d) diverges

e) diverges

f) converges

g) converges
a) converges conditionally

b) converges conditionally

c) diverges

d) converges absolutely

e) diverges

f) converges absolutely

g) diverges

[edit] Full solutions

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

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

d) solution to come

e) solution to come

f) solution to come

g) solution to come
a) solution to come

b) solution to come

c) solution to come

d) solution to come

e) solution to come

f) solution to come

g) solution to come