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Determine whether the series $\sum_{n=1}^{\infty} \frac{\ln (n)}{n}$ converges or diverges.

$P \text { - Series } \Rightarrow \sum_{n=1}^{\infty} \frac{1}{n^{p}}$

(1) $\sum_{n=1}^{\infty} \frac{1}{n^{p}}$ is Convergent if $p>1$

(2) $\sum_{n=1}^{\infty} \frac{1}{n^{p}}$ is divergent if $p \le 1$

$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}=\sum_{n=1}^{\infty} \frac{1}{n^{\frac{1}{2}}} \quad p=\frac{1}{2}<1$ divergent

$\sum_{n=1}^{\infty} \frac{1}{n^{3}}$ Convergent  p = 3

$f(x)=\frac{\ln (x)}{x}$

(1) $f(x) \text { is +ve on }[1, \infty)$

(2) $f(x)$ is Continuous on $[1, \infty)$ because the logarathmic function is continuous

(3) Check $f^{\prime}(x)$

$f^{\prime}(x)=\frac{\left(\frac{1}{x}\right)(x)-(1)(\ln (x))}{x^{2}}=\frac{1-\ln (x)}{x^{2}}$

$1-\ln (x)<0 \Rightarrow \ln (x)>1$

$e^{\ln (x)}>e^{1} \Rightarrow x>e$

f is decreasing for $x>e$

$\int_{1}^{\infty} f(x) d x=\int_{1}^{\infty} \frac{\ln (x)}{x} d x=\lim _{t \rightarrow \infty} \int_{1}^{t} \frac{\ln (x)}{x} d x$

$\int_{1}^{\infty} f(x) d x=\lim _{t \rightarrow \infty} \int_{1}^{t} \frac{\ln (x)}{x} d x=\lim _{t \rightarrow \infty} \int_{1}^{t} \ln (x) \cdot \frac{1}{x} d x$

$\int_{1}^{\infty} f(x) d x=\lim _{t \rightarrow \infty}\left.\frac{(\ln (x))^{2}}{2}\right|_{1} ^{t}=\lim _{t \rightarrow \infty}\left(\frac{(\ln (t))^{2}}{2}-\frac{\left(\ln (1)\right)^{2}}{2}\right)$

$\ln (1) = 0$

$\frac{\left(\ln (1) \right)^{2}}{2} = 0$

$\int_{1}^{\infty} f(x) d x=\lim _{t \rightarrow \infty}\left(\frac{(\ln (t))^{2}}{2}\right)=\frac{(\ln (\infty))^{2}}{2}=\frac{\infty}{2}=\infty$

$\int_{1}^{\infty} \frac{\ln (x)}{x} d x$ is divergent

$\Rightarrow \sum_{n=1}^{\infty} \frac{\ln (n)}{n}$ is divergent too.

Determine whether the series $\sum_{n=1}^{\infty} n \mathrm{e}^{-n^{2}}$ converges or diverges

$f(x)=x e^{-x^{2}}$

(1) $f(x) \text { is +ve on }[1, \infty)$

(2) $f(x)$ is Continuons on $[1, \infty)$

(3) check $f^{\prime}(x)$ for decreasing

$f^{\prime}(x)=(1)\left(e^{-x^{2}}\right)+-2 x e^{-x^{2}}(x)$

$=e^{-x^{2}}-2 x^{2} e^{-x^{2}}=\left(1-2 x^{2}\right) e^{-x^{2}}$

$e^{-x^{2}}>0$ always

$2 x^{2}>1$ always for $x \in[1, \infty)$

$1-2 x^{2}<0$

$f^{\prime}(x)=\left(1-2 x^{2}\right) e^{-x^{2}}<0$

$f(x)$ is decreasing on $[1, \infty)$

$\int_{1}^{\infty} f(x) d x=\int_{1}^{\infty} x e^{-x^{2}} d x=\lim _{t \rightarrow \infty} \int_{1}^{t} x e^{-x^{2}} d x$

Let $u=-x^{2} \Rightarrow d u=-2 x d x$

$\int_{1}^{\infty} f(x) d x=\lim _{t \rightarrow \infty}\left(-\frac{1}{2} \int_{1}^{t}-2 x e^{-x^{2}} d x\right)$

When $x=1 \Rightarrow u=-1$ When $x=t \Rightarrow u=-t^{2}$

$\int_{1}^{\infty} f(x) d x=\lim _{t \rightarrow \infty}\left(-\frac{1}{2} \int_{-1}^{-t^2} e^{u} d u\right)$

$=-\frac{1}{2} \lim _{t \rightarrow \infty}\left(e^{u} \right) |_{-1}^{-t^2}$

$\int_{1}^\infty f(x) d x=-\frac{1}{2} \lim _{t \rightarrow \infty}\left(e^{-t^{2}}-e^{-1}\right)=-\frac{1}{2}\left(e^{-\infty}-e^{-1}\right)$

$e^{-\infty}=0$

$=\frac{e^{-1}}{2}$

$\sum_{n=1}^{\infty} n e^{-n^{2}}$ is convergent by (I.T)