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Question Number 127778 by Bird last updated on 02/Jan/21
$${study}\:{tbe}\:{convergence}\:{of} \\ $$$$\sum_{{n}} ^{\infty} \frac{\mathrm{1}}{{nln}\left(\mathrm{1}+{n}\right)} \\ $$
Answered by mnjuly1970 last updated on 02/Jan/21
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{nln}\left({n}+\mathrm{1}\right)}\:\geqslant\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right){ln}\left({n}+\mathrm{1}\right)} \\ $$$$=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{nln}\left({n}\right)}\:\underset{{theorem}} {\overset{{cauchy}\:{density}} {\approx}}\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}^{{k}} }{\mathrm{2}^{{k}} {ln}\left(\mathrm{2}^{{k}} \right)} \\ $$$$=\:\frac{\mathrm{1}}{{log}\left(\mathrm{2}\right)}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}}\:\rightarrow\infty \\ $$$$\:{harmonic}\:{series}\:\:{is}\:{divergent}.. \\ $$$$\:{then}\:\:{the}\:{orginal}\:{seies}\:\:{is}\:{divergent}... \\ $$$${comparison}\:{test}... \\ $$
Answered by mindispower last updated on 02/Jan/21
![f:x→xln(1+x) positiv increase function ∫_1 ^∞ f(x)dx and Σ(1/(nln(1+n))) sam nature ∫_1 ^∞ (1/(xln(1+x)))dx= By part [((ln(x))/(ln(1+x)))]_1 ^∞ −∫_1 ^∞ ((ln(x))/((1+x)ln^2 (1+x))) when x→∞ ((ln(x))/((1+x)ln^2 (1+x)))∼(1/((1+x)ln(1+x))) x→(1/((1+x)ln(1+x))) not cv in+∞ ∫(dx/((1+x)ln(1+x)))=ln(ln(1+x)→∞ ⇒Σ(1/(nln(1+n))) Dv](Q127823.png)
$${f}:{x}\rightarrow{xln}\left(\mathrm{1}+{x}\right)\:{positiv}\:{increase}\:{function} \\ $$$$\int_{\mathrm{1}} ^{\infty} {f}\left({x}\right){dx}\:{and}\:\Sigma\frac{\mathrm{1}}{{nln}\left(\mathrm{1}+{n}\right)}\:{sam}\:{nature} \\ $$$$\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{xln}\left(\mathrm{1}+{x}\right)}{dx}=\:{By}\:{part}\:\left[\frac{{ln}\left({x}\right)}{{ln}\left(\mathrm{1}+{x}\right)}\right]_{\mathrm{1}} ^{\infty} −\int_{\mathrm{1}} ^{\infty} \frac{{ln}\left({x}\right)}{\left(\mathrm{1}+{x}\right){ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)} \\ $$$${when}\:{x}\rightarrow\infty \\ $$$$\frac{{ln}\left({x}\right)}{\left(\mathrm{1}+{x}\right){ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)}\sim\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right){ln}\left(\mathrm{1}+{x}\right)} \\ $$$${x}\rightarrow\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right){ln}\left(\mathrm{1}+{x}\right)}\:{not}\:{cv}\:{in}+\infty \\ $$$$\int\frac{{dx}}{\left(\mathrm{1}+{x}\right){ln}\left(\mathrm{1}+{x}\right)}={ln}\left({ln}\left(\mathrm{1}+{x}\right)\rightarrow\infty\right. \\ $$$$\Rightarrow\Sigma\frac{\mathrm{1}}{{nln}\left(\mathrm{1}+{n}\right)}\:{Dv} \\ $$$$ \\ $$