Series

Summable Sequence

The sequence $\{a_n\}$ is called summable if the sequence $\{s_n\}$ converges, where
MATH
In this case, $\lim \ s_n$ is denoted by MATH and is called the sum of the sequence $\{a_n\}.$

Cauchy Criterion

The sequence $\{a_n\}$ is summable if and only if
MATH
Therefore, if $\{a_n\}$ is summable then MATH

Divergence of MATH

Grouping the series as
MATH
we see that the sum of the terms in each ( ) is $\geq \frac 12$; thus $\{s_n\}$ is unbounded. Therefore the series $\{\frac 1n\}$ is not summable.

Geometric Series

The infinite series MATH is called the geometric series. If $|x|>1$ then each individual term does not converge to $0$ so the series diverges. If $|x|<1$, then it follows from the two equations
MATH

MATH
that
MATH
or
MATH
Since MATH, it follows that
MATH

Convergence Criterion for Nonnegative Series

Boundedness Criterion

A nonnegative series $\{a_n\}$ is convergent if and only if the set of partial sums $s_n$ is bounded.

Comparison Test

1. Suppose that
MATH
Then if MATH converges, so does MATH

Proof. If
MATH

MATH
then
MATH

Since MATH converges, $\{t_n\}$ is bounded. Therefore $\{s_n\}$ is bounded; consequently MATH is convergent.

2. If $a_n$, $b_n>0$ and MATH, then MATH converges if and only if MATH converges.

Proof. Suppose that MATH converges. Since MATH, there exists some $N$ such that
MATH

Since the series MATH converges, it follows that the series MATH converges and this implies the convergence of the whole series MATH which has only finitely many additional terms.

The converse follows at once since MATH

The Ratio Test

Let $a_n>0$ for all $n$, and suppose that
MATH

(a) If $r<1$ then MATH converges.

(b) If $r>1$ then the terms of $a_n$ do not approach $0$ so MATH diverges.

Proof. (a) Suppose that $r<1$. Fix $s$ with $r<s<1$. The assumption MATH implies that there exists some $N$ such that
MATH
i.e.,
MATH
Thus
MATH

MATH

MATH

MATH

MATH

MATH

Since MATH converges, it follows that MATH converges. Therefore the whole series MATH converges.

(b) Suppose that $r>1$. Fix $s$ with $1<s<r$. Then there exists some $N$ such that
MATH
which means
MATH
This shows that the individual terms of $\{a_n\}$ do not approach $0$, so $\{a_n\}$ is not summable.

The Root Test

Let $a_n\geq 0$ for all $n$, and suppose that
MATH

(a) If $r<1$ then MATH converges.

(b) If $r>1$ then the terms of $a_n$ do not approach $0$ so MATH diverges.

Proof. (a) Suppose that $r<1$. Fix $s$ with $r<s<1$. The assumption
MATH
implies that there is some $N$ such that
MATH
i.e.,
MATH
Thus
MATH

MATH

MATH

MATH

MATH
Since the geometric series MATH converges, it follows that MATH converges. Therefore the whole series MATH converges.

(b) Suppose that $r>1$. Fix $s$ with $1<s<r$. Then there exists some $N$ such that
MATH
which means
MATH

This shows that the individual terms of $\{a_n\}$ do not approach $0$, so $\{a_n\}$ is not summable.

Convergence of the Exponential Series MATH

Let $x$ be any positive number. The ratio
MATH
converges to $0$ as $n$ approaches infinity. Hence it follows from the ratio test that the series MATH converges for any positive number $x$. Consequently we have
MATH

Improper Integral

If the limit MATH exists, it is denoted by MATH, and called an improper integral.

The improper integral MATH is defined as MATH. If both improper integrals MATH and MATH exist, then the improper integral MATH exists and equals MATH .

Note that MATH does not exist although the limit MATH exists.

There is another kind of improper integral in which the interval is bounded, but the function is unbounded. If $f$ is an unbounded function such that for any $\epsilon $ with MATH is bounded on $[a+\epsilon ,b]$ and the limit
MATH
exists, then the improper integral $\int_a^bf(x)dx$ is said to exist and is equal to MATH Similarly, if $f$ is an unbounded function such that for any $\epsilon $ with MATH is bounded on $[a,b-\epsilon ]$ and the limit
MATH
exists, then the improper integral $\int_a^bf(x)dx$ is said to exist and is equal to MATH

Integral Test

Suppose that $f$ is positive and decreasing on $[1,\infty )$, and that $f(n)=a_n$ for all $n$. Then MATH converges if and only if the improper integral
MATH

Proof. The convergence of MATH is equivalent to the convergence of the series
MATH
Since $f$ is decreasing we have
MATH
Therefore, if MATH exists, then the series MATH converges and so it follows from the comparison test that MATH converges, and so MATH converges.

If MATH converges, the second half of the inequality shows that the series MATH converges and so the improper integral MATH exists.

Exercise

Establish convergence or divergence by a comparison test

1. MATH 2. MATH 3. MATH 4. MATH 5. MATH

6. MATH 7. MATH 8. MATH 9. MATH 10. MATH

Decide for or against convergence based on the integral test

11. MATH 12. MATH 13. MATH 14. MATH

15. MATH 16. MATH 17. MATH 18. MATH

Decide convergence and name your test

19. MATH 20. MATH 21. MATH 22. MATH 23. MATH

24. MATH 25. MATH(test all $p$) 26. MATH(test all $p$)

27. MATH28. MATH(test all $p,q$)

29. (a) Show that MATH

(b) Express MATH as a Riemann sum and show that MATH approaches $\ln 2.$

30. Compute the sum of the telescoping series
MATH

MATH

31. Compute the sum
MATH
by considering the Taylor expansion of $\tan ^{-1}x.$

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