Solutions - 0180-05

(a)

Set $f(x)={\displaystyle \frac{\ln x}{x^2}}$. Applicability of the IT:

• $f(x)$ is is continuous other than at $x=0$, and the series starts at $1$.
• $f(x)>0$ since $\ln x>0$ for all $x>1$, and $x^2>0$ for all $x\ge 1$.
• $f^{\prime }(x)={\displaystyle \frac{x-2x\ln x}{x^3}}$. This is zero at $x=\sqrt{e}$. For $x>\sqrt{e}$, $f^{\prime }(x)<0$, so the function is decreasing.

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Apply the integral test:

$$\begin{array}{c}{\int }_2^{\infty }\frac{\ln x}{x^2}dx=\underset{R\to \infty }{\lim }{\int }_2^R\frac{\ln x}{x^2}dx\\ \\ \gg \gg \underset{R\to \infty }{\lim }{[-\frac{\ln x}{x}-\frac{1}{x}]|}_2^R\\ \\ \gg \gg (0-0)-(-\frac{\ln 2}{2}--\frac{1}{2})\gg \gg \frac{1+\ln 2}{2}\end{array}$$

This is finite, so the original series converges by the IT.

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(b) For very large $n$, the large powers dwarf the small powers, and the terms look like ${\displaystyle \frac{n^3}{n^5}}$ which equals ${\displaystyle \frac{1}{n^2}}$. So we take this for a comparison series and apply the DCT:

$$\frac{n^3}{n^5+4n+1}<\frac{n^3}{n^5}=\frac{1}{n^2}$$

But ${\displaystyle \sum \frac{1}{n^2}}$ converges ($p=2>1$). So by the DCT, the original series converges.

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(c) For very large $n$, the large powers dwarf the small powers, and the terms look like ${\displaystyle \frac{n^2}{n^4}}$ which equals ${\displaystyle \frac{1}{n^2}}$. So we take this for a comparison series and apply the LCT:

$$a_n\cdot b_n^{-1}=\frac{n^2}{n^4-1}\cdot \frac{n^2}{1}\gg \gg \frac{n^4}{n^4-1}\stackrel{n\to \infty }{⟶}1=L$$

Observe that $0<L<\infty$. Therefore the LCT says that both series converge or both diverge.

We know that ${\displaystyle \sum \frac{1}{n^2}}$ converges ($p=2>1$). So by the LCT, the original series converges.