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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 \(x=t \Rightarrow u=-t^{2}\)

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

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

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



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

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