Tag: Series

Test convergence of the series with a_n = 1/n if n odd and a_n = 1/n² if n even

by RoRi

March 29, 2016

Test the series $\sum_{n=1}^{\infty} a_n$ for convergence where the terms $a_n$ are defined by

$a_n = \begin{cases} \frac{1}{n} & \text{if } n \text{ is odd}, \\ \frac{1}{n^2} & \text{if } n \text{ is even}. \end{cases}$

The series diverges.

Proof. (This proof might be overkill. You could simply observe that this series is greater than the sum of the reciprocals of the odd integers, and use limit comparison with $\frac{1}{n}$ .)Using the piecewise definition of $a_n$ we have

$\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \left( \frac{1}{2n-1} + \frac{1}{(2n)^2} \right) = \sum_{n=1}^{\infty} \frac{4n^2 + 2n - 1}{8n^3 - 4n^2}.$

Then using the limit comparison we have,

$\begin{align*} \lim_{n \to \infty} \frac{ \frac{4n^2 + 2n - 1}{8n^3 - 4n^2}}{\frac{1}{n}} &= \lim_{n \to \infty} \frac{4n^3 + 2n^2 - n}{8n^3 - 4n^2} \\[9pt] &= \frac{1}{2}. \end{align*}$

Since we know $\sum \frac{1}{n}$ diverges, the limit comparison test establishes the divergence of

$\sum a_n. \qquad \blacksquare$

Determine whether ∑ 1 / n^{1 + 1/n} converges

by RoRi

March 29, 2016

Test the following series for convergence:

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

The series diverges.

Proof. First, we write

$\frac{1}{n^{1+\frac{1}{n}}} = \frac{1}{n \cdot n^{\frac{1}{n}}}.$

Then, we know $n < 2^n$ for all $n \geq 1$ . (We can deduce this from the Bernoulli inequality, $(1+x)^n \geq 1 + nx$ with $x = 1$ . We proved the Bernoulli inequality in this exercise, Section I.4.10, Exercise #14.) Therefore, $n^{\frac{1}{n}} < 2$ and we have

$\frac{1}{n \cdot n^{\frac{1}{n}}} > \frac{1}{2n}.$

Since the series $\sum \frac{1}{2n}$ diverges we have established the divergence of the given series

$\sum_{n=1}^{\infty} \frac{1}{n^{1+\frac{1}{n}}}. \qquad \blacksquare$

Determine whether ∑ 1 / (log n)^{log n} converges

by RoRi

March 29, 2016

Test the following series for convergence:

$\sum_{n=2}^{\infty} \frac{1}{(\log n)^{\log n}}.$

The series converges.

Proof. First, we use the definition of the exponential to rewrite the terms of the series,

$\frac{1}{(\log n)^{\log n}} = \frac{1}{e^{(\log n)(\log (\log n))}} = \frac{1}{n^{\log (\log n)}}.$

But then for sufficiently large $n$ we know $\log (\log n) > 2$ (since $\log (\log n)$ is strictly increasing and unbounded above). Therefore, for all sufficiently large $n$ we have $\frac{1}{(\log n)^{\log n}} < \frac{1}{n^2}$ ; hence, by the comparison test the series

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

converges $. \qquad \blacksquare$

Determine whether ∑ n^s((n+1)^1/2 – 2n^1/2 + (n-1)^1/2) converges

by RoRi

March 29, 2016

Test the following series for convergence:

$\sum_{n=1}^{\infty} n^s \left( \sqrt{n+1} - 2 \sqrt{n} + \sqrt{n-1} \right).$

The series is convergent if $s < \frac{1}{2}$ and divergent if $s \geq \frac{1}{2}$ .

Proof. We will use the limit comparison test with series $\sum \frac{1}{n^{\frac{3}{2} - s}}$ (which we know converges for $s < \frac{1}{2}$ and diverges for $s \geq \frac{1}{2}$ ). First, we rewrite the terms in the series,

$\begin{align*} \left|n^s \left( \sqrt{n+1} - 2 \sqrt{n} + \sqrt{n-1} \right) \right|&= n^s \left| \left( \frac{ \left( \sqrt{n+1} - 2\sqrt{n} + \sqrt{n-1}\right)\left(\sqrt{n+1} + 2 \sqrt{n} + \sqrt{n-1} \right)}{ \sqrt{n+1} + 2\sqrt{n} + \sqrt{n-1}} \right) \right|\\[9pt] &= n^s \left| \left( \frac{2 \sqrt{n+1}\sqrt{n-1} - 2n}{\sqrt{n+1} + 2\sqrt{n} + \sqrt{n-1}} \right) \right| \\[9pt] &= n^s \left| \left( \frac{2 \left( \sqrt{n^2 - 1} - n \right)}{2 \sqrt{n} + \sqrt{n+1} + \sqrt{n-1}} \right) \right|\\[9pt] &= n^s \left| \left( \frac{2 \left( \sqrt{n^2 - 1} - n \right) \left( \sqrt{n^2-1} + n \right)}{\left( 2 \sqrt{n} + \sqrt{n+1} + \sqrt{n-1} \right) \left( \sqrt{n^2-1} + n \right)} \right) \right|\\[9pt] &= n^s \left|\left( \frac{-2}{\left( 2 \sqrt{n} + \sqrt{n+1} + \sqrt{n-1} \right) \left( \sqrt{n^2-1} + n \right)} \right)\right| \\[9pt] &= n^2 \left( \frac{2}{ \left( 2 \sqrt{n} + \sqrt{n+1} + \sqrt{n-1} \right) \left( \sqrt{n^2-1} + n \right)} \right). \end{align*}$

Now, we can use the limit comparison test with $\frac{1}{n^{\frac{3}{2} - s}}$ .

$\begin{align*} & \lim_{n \to \infty} \frac{ n^s \left( \frac{2}{\left( 2 \sqrt{n} + \sqrt{n+1} + \sqrt{n-1} \right) \left( \sqrt{n^2-1} + n \right)} \right) }{ \frac{1}{n^{\frac{3}{2}-s}} } \\[9pt] &= \lim_{n \to \infty} n^{\frac{3}{2}} \left( \frac{2}{\left( 2 \sqrt{n} + \sqrt{n+1} + \sqrt{n-1} \right) \left( \sqrt{n^2-1} + n \right)} \right) \\[9pt] &= \lim_{n \to \infty} \frac{2}{\left( 2 + \sqrt{1 + \frac{1}{n}} + \sqrt{1 - \frac{1}{n}} \right) \left( \sqrt{1 - \frac{1}{n^2}} + 1 \right)} \\[9pt] &= \frac{1}{4}. \end{align*}$

(In the second to last line, we multiplied the numerator and denominator by $n^{-\frac{3}{2}}$ and in the denominator we had $n^{-\frac{3}{2}} = n^{-\frac{1}{2}}\cdot n^{-1}$ and split these two pieces between the two terms in the product of the denominator.) Therefore, the given series converges or diverges as $\sum \frac{1}{n^{s-\frac{3}{2}}}$ does. Since $\sum \frac{1}{n^{s - \frac{3}{2}}}$ converges for $s < \frac{1}{2}$ and diverges for $s \geq \frac{1}{2}$ , we have proved our claim $. \qquad \blacksquare$

Determine whether the series ∑ ((1 + n²)^1/2 – n) converges

by RoRi

March 29, 2016

Test whether the following series converges:

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

This series diverges.

Proof. We use the comparison test with the known divergent series $\sum \frac{1}{3n}$ . First,

$\begin{align*} \left( n + \frac{1}{3n} \right)^2 &= n^2 + \frac{2}{3} + \frac{1}{9n^2} \\[9pt] &< n^2 + 1 \end{align*}$

since $\frac{1}{9n^2} < \frac{1}{3}$ for all $n \geq 1$ . Therefore,

$\begin{align*} && n^2 + 1 &> \left( n + \frac{1}{3n} \right)^2 \\[9pt] \implies && \sqrt{n^2 + 1} &> n + \frac{1}{3n} \\[9pt] \implies && \sqrt{n^2 + 1} - n &> \frac{1}{3n}. \end{align*}$

Thus, by the comparison test we have established the divergence of

$\sum_{n=1}^{\infty} \left( \sqrt{1+n^2} - n \right). \qquad \blacksquare$

Prove some properties of the sequence e^a_n = a_n + e^b_n

by RoRi

March 29, 2016

Let $\{ a_n \}$ and $\{ b_n \}$ be two sequences that satisfy the relationship

$e^{a_n} = a_n + e^{b_n}$

for all $n$ .

Prove that if $a_n > 0$ for all $n$ then $b_n > 0$ for all $n$ .
Prove that if $a_n > 0$ for all $n$ and if $\sum a_n$ is convergent, then
$\sum \frac{b_n}{a_n} \qquad \text{converges}.$

Proof. Assume $a_n > 0$ for all $n$ . From the given relation between the $a_n$ and $b_n$ we have
$e^{a_n} = a_n + e^{b_n} \quad \implies \quad e^{b_n} = e^{a_n} - a_n.$

Now, we claim $e^{a_n} - a_n > 1$ for all $n$ . To see this, consider the function

$f(x) = e^x - x \qquad \implies \qquad f'(x) = e^x - 1.$

Since $f'(x) > 0$ for all $x > 0$ we have $f(x)$ is increasing for all $x > 0$ . Since $f(0) = 1$ we must have $f(x) > 1$ for all $x > 0$ . Since $a_n > 0$ by assumption we then have

$e^{a_n} - a_n > 1 \qquad \text{for all } n.$

But, this implies

$e^{b_n} > 1 \quad \implies \quad b_n > 0$

for all $n. \qquad \blacksquare$
Proof. To prove $\sum \frac{b_n}{a_n}$ converges we use the limit comparison test with $\sum a_n$ . First, since $\sum a_n$ converges we know
$\lim_{n \to \infty} a_n = 0.$

Now, we use the given relation between the $a_n$ and the $b_n$ ,

$e^{a_n} = a_n + e^{b_n} \quad \implies \quad b_n = \log \left( e^{a_n} - a_n \right) = \log \left( 1 + \left( e^{a_n} - 1 - a_n \right) \right).$

Then, use the expansion of the exponential,

$e^{a_n} = 1 + a_n + \frac{a_n^2}{2} + \cdots, \quad \implies \quad e^{a_n} - 1 - a_n = \frac{a_n^2}{2} + \frac{a_n^3}{6} + \cdots.$

This gives us

$b_n = \frac{a_n^2}{2} + \frac{a_n^3}{6} + \cdots.$

Therefore,

$\begin{align*} \lim_{n \to \infty} \frac{\frac{b_n}{a_n}}{a_n} &= \lim_{n \to \infty} \frac{b_n}{a_n^2} \\[9pt] &= \lim_{n \to \infty} \frac{ \frac{a_n^2}{2} + \frac{a_n^3}{6} + \cdots }{a_n^2} \\[9pt] &= \frac{1}{2} + \lim_{n \to \infty} \left( \frac{a_n}{6} + \cdots \right) \\[9pt] &= \frac{1}{2}. &(\lim_{n \to \infty} a_n = 0.) \end{align*}$

Therefore, by the limit comparison test the series $\sum \frac{b_n}{a_n}$ and $\sum a_n$ either both converge or both diverge. Since $\sum a_n$ converges by hypothesis, we have established the convergence of $\sum \frac{b_n}{a_n}. \qquad\blacksquare$

Prove or disprove two statements about absolute convergence

by RoRi

March 18, 2016

Prove or disprove the following two statements.

If $\sum a_n$ is absolutely convergent, then
$\sum \frac{a_n^2}{(1+a_n^2)}$

is absolutely convergent.
If $\sum a_n$ is absolutely convergent, and if $a_n \neq -1$ for all $n$ , then,
$\sum \frac{a_n}{1+a_n}$

is absolutely convergent.

Proof. Since $\sum a_n$ converges absolutely, we know $\sum |a_n|$ converges. By a previous exercise (Section 10.20, Exercise #50) we know that this implies that $\sum a_n^2$ converges. Then we have,
$\frac{a_n^2}{1+a_n^2} \leq a_n^2$

for all $n$ since $1+ a_n^2 \geq 1$ . Hence, by comparison with $\sum a_n^2$ we have the convergence of

$\sum_{n=1}^{\infty} \frac{a_n^2}{1+a_n^2}. \qquad \blacksquare$
Proof. Since $\sum a_n$ converges absolutely, we know $\sum |a_n|$ converges. Then, using the limit comparison test we have,
$\begin{align*} \lim_{n \to \infty} \frac{|a_n|}{\left| \frac{a_n}{1+a_n} \right|} &= \lim_{n \to \infty} |1 + a_n| \\[9pt] &= 1. \end{align*}$

The final equality follows since $\sum a_n$ converges means we must have $\lim_{n \to \infty} a_n = 0$ . Hence, the series $\sum \left| \frac{a_n}{1+a_n} \right|$ converges; and thus,

$\sum \frac{a_n}{1+a_n}$

converges absolutely (as long as $a_n \neq -1$ so that all of the terms are defined) $. \qquad \blacksquare$

Prove that if ∑ a_n converges then ∑ (a_n)^1/2 n^-p converges for p > 1/2

by RoRi

March 18, 2016

Let $\sum a_n$ be a convergent series of nonnegative terms (i.e., $a_n \geq 0$ for all $n$ ). Prove that

$\sum \sqrt{a_n} n^{-p}$

converges for all $p > \frac{1}{2}$ . Provide an example to show this fails in the case $p = \frac{1}{2}$ .

(Note. Thanks to Giovanni in the comments for pointing out the problems with the original proof.)

Proof. We apply the Cauchy-Schwarz inequality. (See this exercise for some more proofs regarding the Cauchy-Schwarz inequality.) The Cauchy-Schwarz inequality establishes that

$\left( \sum_{n=1}^{N} b_n c_n \right)^2 \leq \left( \sum_{n=1}^{N} b_n^2 \right) \left( \sum_{n=1}^{N} c_n^2 \right).$

So, to apply this to the current problem, we let

$b_n = \sqrt{a_n}, \qquad c_n = \frac{1}{n^p}.$

Then, since $a_n \geq 0$ we have $a_n = |a_n|$ ; hence, $\left(\sqrt{a_n}\right)^2 = a_n$ . So, we have

$\begin{align*} && \left( \sum_{n=1}^{N} b_n c_n \right)^2 &\leq \left( \sum_{n=1}^{N} b_n^2 \right) \left( \sum_{n=1}^{N} c_n^2 \right) \\[9pt] \implies && \left( \sum_{n=1}^{N} \frac{\sqrt{a_n}}{n^p} \right)^2 &\leq \left( \sum_{n=1}^{N} a_n \right) \left( \sum_{n=1}^{N} \frac{1}{n^{2p}} \right) \\[9pt] \implies && \left( \sum_{n=1}^{N} \frac{\sqrt{a_n}}{n^p} \right) & \leq \left( \sum_{n=1}^{N} a_n \right)^{\frac{1}{2}} \right) \left( \sum_{n=1}^{N} \frac{1}{n^{2p}} \right)^{\frac{1}{2}} \\[9pt] \implies && \left( \sum_{n=1}^{N} \frac{\sqrt{a_n}}{n^p} \right) &\leq \left( \sum_{n=1}^{\infty} a_n \right)^{\frac{1}{2}} \left( \sum_{n=1}^{\infty} \frac{1}{n^{2p}} \right)^{\frac{1}{2}}. \end{align*}$

But, on the right we know $\sum a_n$ converges (by hypothesis) and we know $\sum \frac{1}{n^{2p}}$ converges for $p > \frac{1}{2}$ . Hence, each of the sums on the right converges, so the term on the right is some finite number $M$ . This means the partial sums of $\sum \frac{\sqrt{a_n}}{n^p}$ are bounded by a constant $M$ . By Theorem 10.7 (page 395 of Apostol) this implies the convergence of the series

$\sum_{n=1}^{\infty} \frac{\sqrt{a_n}}{n^p}. \qquad \blacksquare$

Example. Let $p = \frac{1}{2}$ and let $a_n = \frac{1}{n \log^2 n}$ . Then, we know $\sum a_n$ converges (by Example #2 on page 398 of Apostol, established with the integral test). However,

$\sqrt{a_n} = \sqrt{n} \log n \quad \implies \quad \sqrt{a_n} \cdot n^{-\frac{1}{2}} = \frac{1}{n \log n}.$

We know

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

diverges by the same example on page 398. Therefore, we conclude the theorem fails in the case $p = \frac{1}{2}$ .

If ∑ |a_n| converges prove ∑ a_n² converges

by RoRi

March 17, 2016

Prove that if the series $\sum |a_n|$ converges then the series $\sum a_n^2$ also converges. Also, give an example to show that the converse is false, i.e., a case in which $\sum a_n^2$ converges but $\sum |a_n|$ does not.

Proof. Assume $\sum |a_n|$ converges. Then we know $\lim_{n \to \infty} |a_n| = 0$ . Thus, by the definition of the limit, for all $\varepsilon > 0$ there exists an integer $N$ such that

$|a_n| < \varepsilon \qquad \text{whenever} \qquad n \geq N.$

Taking $\varepsilon = 1$ we then know there exists an $N$ such that

$|a_n| < 1 \qquad \text{whenever} \qquad n \geq N.$

But then, $a_n^2 = |a_n|^2 < |a_n|$ if $n \geq N$ . Thus, we have

$\sum_{n=1}^{\infty} a_n^2 = \sum_{n=1}^N a_n^2 + \sum_{n=N+1}^{\infty} a_n^2.$

The first sum is a finite sum so it converges, and the second sum converges by comparison with $\sum |a_n|$ . Hence, $\sum a_n^2$ converges $. \qquad \blacksquare$

Counterexample. The converse is false. Let $a_n = \frac{1}{n}$ . Then,

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

converges. However,

$\sum_{n=1}^{\infty} |a_n| = \sum_{n=1}^{\infty} \left| \frac{1}{n} \right| = \sum_{n=1}^{\infty} \frac{1}{n}$

diverges.

Prove that ∑ 1/a_n diverges if ∑ a_n converges

by RoRi

March 17, 2016

Prove that if $a_n > 0$ for all $n \in \mathbb{Z}$ and if $\sum_n a_n$ converges, then $\sum_n \frac{1}{a_n}$ diverges.

Proof. Since $\sum a_n$ converges we know $\lim_{n \to \infty} a_n =0$ . By the definition limit this means that for all $\varepsilon > 0$ there exists an integer $N$ such that

$|a_n| < \varepsilon \qquad \text{whenever} \qquad n \geq N.$

In particular, if we take $\varepsilon = 1$ then we know there is some $N$ such that $n \geq N$ implies $|a_n| < 1$ . But then,

$|a_n| < 1 \quad \implies \quad \left| \frac{1}{a_n} \right| > 1 \quad \implies \quad \lim_{n \to \infty} \frac{1}{a_n} \neq 0.$

Hence, the series $\sum \frac{1}{a_n}$ diverges $. \qquad \blacksquare$