$$\gdef\diff#1#2{\frac{\mathrm{d}#1}{\mathrm{d}#2}} \gdef\dydx{\diff{y}{x}} \gdef\dd{\mathop{}\!\mathrm{d}} \gdef\nimplies{\;\;\;\not\nobreak\!\!\!\!\implies\;} \gdef\({\left(} \gdef\){\right)} \gdef\[{\left[} \gdef\]{\right]} \gdef\R{\mathbb{R}} \gdef\arraystretch{2.2em}$$

Homework 9

Mos Kullathon
921425216

Determine if the following series is convergent or divergent. You can use any tests you have learnt so far.

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

$\displaystyle\lim_{n\to\infty}\sin\(n\frac{\pi}{2}\)$ does not exist. By divergence test, the series diverges.

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

Let $\displaystyle f(x) = \frac{1}{x\ln^2x}$.

Since $\displaystyle f(x)=\frac{1}{x\ln^2x}>0$ and $x\ln^2x$ is monotonically increasing for $x\ge2$, so its reciprocal $\displaystyle f(x) = \frac{1}{x\ln^2x}$ is decreasing for $x\ge2$.

By integral test:
$$\int_2^\infty\frac{1}{x(\ln x)^2}\dd x = \lim_{n\to\infty}\int_2^n \frac{1}{x(\ln x)^2}\dd x \\ \begin{darray}{cc} u = \ln x &\implies& \dd u = \frac{1}{x}\dd x \\ &\iff& \dd x = x\dd u \\ x = 2 &\implies& u = \ln 2 \\ x = n &\implies& u = \ln n \end{darray} \\ \begin{align*} \lim_{n\to\infty}\int_{\ln2}^{\ln n} \frac{1}{x(\ln x)^2}\dd x &= \lim_{n\to\infty}\int_{\ln2}^{\ln n} \frac{x\dd u}{x u^2} \\ &= \lim_{n\to\infty}\int_{\ln2}^{\ln n} u^{-2}\dd u \\ &= \lim_{n\to\infty}\[-\frac{1}{u}\]_{\ln2}^{\ln n} \\ &= \lim_{n\to\infty}-\frac{1}{\ln n} + \frac{1}{\ln 2} \\ &= \frac{1}{\ln 2} \end{align*}$$

Since $\displaystyle\int_2^\infty\frac{1}{x(\ln x)^2}\dd x = \frac{1}{\ln 2} < \infty$, the series is convergent.

(c) $\sum_{n=1}^\infty \frac{1}{n^{3/2}+1}$

Let $\displaystyle a_n = \frac{1}{n^{3/2}+1}$ and $\displaystyle b_n = \frac{1}{n^{3/2}}$.

Then,
$$\frac{a_n}{b_n} = \frac{n^{3/2}}{n^{3/2}+1}. \\ \therefore \lim_{n\to\infty}\frac{a_n}{b_n}=1$$

By limit comparison test, $\displaystyle\sum_{n=1}^\infty a_n$ and $\displaystyle\sum_{n=1}^\infty b_n$ will either converge or diverge together.

Let $\displaystyle f(x)=\frac{1}{x^{3/2}}$.

Since $\displaystyle f(x)>0$ and $\displaystyle f'(x)=-\frac{3}{2x^{5/2}} < 0$ for all $x\ge1$, we apply the integral test.

$$\begin{align*} \int_1^\infty\frac{1}{x^{3/2}} \dd x &= \lim_{n\to\infty}\int_1^n x^{-3/2}\dd x \\ &= \lim_{n\to\infty}\[\frac{x^{-1/2}}{-1/2}\]_1^n \\ &= \lim_{n\to\infty}\[-\frac{2}{\sqrt{x}}\]_1^n \\ &= \lim_{n\to\infty}-\frac{2}{\sqrt{n}} + 2 \\ &= 2 \end{align*}$$

By integral test, $\displaystyle \sum_{n=1}^\infty \frac{1}{n^{3/2}}$ is convergent. Subsequently, by the limit comparison test, the series $\displaystyle\sum_{n=1}^\infty \frac{1}{n^{3/2}+1}$ must also be convergent.

(d) $\sum_{n=1}^\infty \frac{\sqrt{n}}{(n+1)^2}$

Let $\displaystyle a_n = \frac{\sqrt{n}}{(n+1)^2} = \frac{n^{1/2}}{n^2+2n+1}$ and $\displaystyle b_n = \frac{1}{n^{3/2}}$.

Then,
$$\frac{a_n}{b_n} = \frac{n^2}{n^2+2n+1}. \\ \therefore \lim_{n\to\infty}\frac{a_n}{b_n}=1$$

Since $\displaystyle b_n = \frac{1}{n^{3/2}}$ is nonnegative and decreasing, we apply the integral test.

From (c), we concluded that $\displaystyle \sum_{n=1}^\infty\frac{1}{n^{3/2}}$ is convergent by integral test. Subsequently, by the limit comparison test, the series $\displaystyle\sum_{n=1}^\infty \frac{\sqrt{n}}{(n+1)^2}$ must also be convergent.

(e) $\sum_{n=2}^\infty \frac{(\ln n)^2}{n}$

Since
$$\ln n = 1 \iff n =e,$$

and
$$\frac{\dd}{\dd n}\ln n = \frac{1}{n} > 0, \forall n>3$$

we have that
$$\ln n > 1 ,\forall n>3.$$

As such, for all values $n>3$:
$$\ln n > 1 \\ \iff \ln^2n > 1 \\ \iff \frac{\ln^2 n}{n} > \frac{1}{n}.$$

Since $\displaystyle\sum_{n=2}^\infty\frac{1}{n}$ is divergent, by comparison test $\displaystyle\sum_{n=2}^\infty\frac{(\ln n)^2}{n}$ must also be divergent.

(f) $\sum_{n=1}^\infty \frac{n^2}{(n^2+10)^2}$

Let $\displaystyle a_n = \frac{n^2}{(n^2+10)^2} = \frac{n^2}{n^4 +20n^2 + 100}$ and $\displaystyle b_n =\frac{1}{n^2}$.

Then,
$$\frac{a_n}{b_n} = \frac{n^4}{n^4+20n^2+100} \\ \therefore\lim_{n\to\infty}\frac{a_n}{b_n} = 1$$

Since $\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}$ converges, then by limit comparison $\displaystyle \sum_{n=1}^\infty \frac{n^2}{(n^2+10)^2}$ also converges.

(g) $\sum_{n=1}^\infty \frac{n\ln n}{n^3+2}$

For sufficiently large $n$,
$$\frac{n\ln n}{n^3+2}\approx\frac{n}{n^3+2}$$

Let $\displaystyle a_n = \frac{n}{n^3+2}$ and $\displaystyle b_n = \frac{n^\epsilon}{n^2}=\frac{1}{n^{2-\epsilon}}$ for some small $\epsilon>0$.

Then,
$$\frac{a_n}{b_n} = \frac{n^{3-\epsilon}}{n^3+2} \\ \therefore \lim_{n\to\infty}\frac{a_n}{b_n}=0$$

For small values of $\epsilon>0$, the series $\displaystyle\sum_{n=1}^\infty\frac{1}{n^{2-\epsilon}}$ converges. As such, by limit comparison test $\displaystyle\sum_{n=1}^\infty \frac{n\ln n}{n^3+2}$ converges.

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

$$\ln e = 1 \implies \ln n > 1 ,\forall n>3$$

So for $n>3$:
$$\ln n > 1 \\ \iff \ln^2 n > 1 \\ \iff n^2\ln^2n > n^2 \\ \iff \frac{1}{n^2\ln^2n} < \frac{1}{n^2}$$

Since $\displaystyle\sum_{n=2}^\infty\frac{1}{n^2}$ converges, by comparison test $\displaystyle\sum_{n=2}^\infty\frac{1}{n^2(\ln n)^2}$ also converges.

Section 5.5

State whether each of the following series converges absolutely, conditionally, or not at all.

(251) $\displaystyle \sum_{n=1}^\infty (-1)^{n+1}\frac{\sqrt{n}+1}{\sqrt{n}+3}$

Let $\displaystyle a_n = \frac{\sqrt{n}+1}{\sqrt{n}+3}$.

$$\lim_{n\to\infty}a_n = \frac{\sqrt{n}+1}{\sqrt{n}+3} = 1 \neq 0$$

By divergence test, the series diverges.

(252) $\displaystyle \sum_{n=1}^\infty (-1)^{n+1}\frac{1}{\sqrt{n+3}}$

Let $\displaystyle a_n = \frac{1}{\sqrt{n+3}}$.

Since $\sqrt{n+3}$ is monotonically increasing, its reciprocal $\displaystyle a_n=\frac{1}{\sqrt{n+3}}$ must subsequently be decreasing.

And
$$\lim_{n\to\infty}a_n =\lim_{n\to\infty}\frac{1}{\sqrt{n+3}} = 0.$$

By alternating series test, $\displaystyle \sum_{n=1}^\infty (-1)^{n+1}\frac{1}{\sqrt{n+3}}$ is convergent.

Testing for absolute convergence

$$\sum_{n=1}^\infty \left|(-1)^{n+1}\frac{1}{\sqrt{n+3}}\right| = \sum_{n=1}^\infty \frac{1}{\sqrt{n+3}}$$

Let $\displaystyle f(x) = \frac{1}{\sqrt{x+3}}$.

We note that $f(x) >0$ and $f'(x)<0$ for $x\in[1,\infty)$.

$$\begin{align*} \int_1^\infty\frac{1}{\sqrt{x+3}}\dd x &= \lim_{n\to\infty}\int_1^n(x+3)^{-1/2}\dd x \\ &= \lim_{n\to\infty}\[2\sqrt{x+3}\]_1^n \\ &= \lim_{n\to\infty} 2\sqrt{n+3}-4 \\ &= \infty \end{align*}$$

By integral test, we find that the sum of the absolute values of the terms do not converge.

As such, we conclude that $\displaystyle \sum_{n=1}^\infty (-1)^{n+1}\frac{1}{\sqrt{n+3}}$ exhibits conditional convergence.

(260) $\displaystyle \sum_{n=1}^\infty (-1)^{n+1}\sin^2(1/n)$

Since $\displaystyle\sin x\le x$ for $x>0$. Then, where $a_n=\sin^2(1/n)$, we have that $0\le a_{n+1}\le a_n$ for $n\ge1$.

$$\lim_{n\to\infty}\sin^2(1/n) =\sin^2\(\lim_{n\to\infty}\frac{1}{n}\) =\sin^20=0$$

By alternating series test, the series is convergent.

Testing for absolute convergence

$$\sum_{n=1}^\infty\left|(-1)^{n+1}\sin^2(1/n)\right| =\sum_{n=1}^\infty \sin^2(1/n)$$

For $n>0$:
$$\sin n \le n \\ \iff \sin\frac{1}{n} \le \frac{1}{n} \\ \iff \sin^2\frac{1}{n} \le \frac{1}{n^2}$$

Since $\displaystyle\sum_{n=1}^\infty\frac{1}{n^2}$ converges, by comparison test $\displaystyle\sum_{n=1}^\infty\sin^2\frac{1}{n}$ also converges. As such, we conclude that $\displaystyle\sum_{n=1}^\infty(-1)^{n+1}\sin^2(1/n)$ exhibits absolute convergence.

(261) $\displaystyle \sum_{n=1}^\infty (-1)^{n+1}\cos^2(1/n)$

$$\lim_{n\to\infty}\cos^2(1/n) =\cos^2\(\lim_{n\to\infty}\frac{1}{n}\) =\cos^20 =1\neq0$$

By divergence test, the series is divergent.