Question Number 96951 by aurpeyz last updated on 05/Jun/20
$${prove}\:{that}\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+…+\frac{−\mathrm{1}^{{n}−\mathrm{1}} }{{n}}\:\:{is}\:{always}\:{positive} \\ $$$$ \\ $$
Commented by mr W last updated on 05/Jun/20
$${how}\:{many}\:{times}\:{do}\:{you}\:{want}\:{to}\:{post} \\ $$$${the}\:{same}\:{question}?\:{besides}\:{the}\: \\ $$$${question}\:{is}\:{already}\:{answered}.\:{if}\:{you} \\ $$$${don}'{t}\:{understand}\:{the}\:{answer},\:{please} \\ $$$${give}\:{feedback}\:{in}\:{the}\:{same}\:{thread}\:{as} \\ $$$${the}\:{original}\:{question}\:{and}\:{answer}, \\ $$$${instead}\:{of}\:{posting}\:{the}\:{same}\:{question} \\ $$$${a}\:{second}\:{or}\:{third}\:{time}.\:{thank}\:{you}! \\ $$
Commented by mr W last updated on 05/Jun/20
$${answer}\:{in}\:{Q}\mathrm{95473} \\ $$
Commented by mr W last updated on 05/Jun/20
$${you}\:{can}\:{click}\:{on}\:{the}\:{funel}\:{icon}\:{on}\:{the} \\ $$$${top}\:{of}\:{the}\:{screen}\:{and}\:{select}\:“{Go}\:{to} \\ $$$${question}\:{ID}''\:{and}\:{then}\:{enter}\:{the}\:{no}. \\ $$$${of}\:{the}\:{question}\:{you}\:{want}\:{to}\:{go}\:{to}. \\ $$$${but}\:{in}\:{this}\:{case}\:{a}\:{command}\:{bar}\:{is} \\ $$$${automatically}\:{generated},\:{just}\:{tap} \\ $$$${on}\:{it}! \\ $$
Commented by aurpeyz last updated on 05/Jun/20
$${i}\:{am}\:{sorry}.\:{iam}\:{new}\:{to}\:{this}\:{platform} \\ $$
Commented by aurpeyz last updated on 05/Jun/20
$${how}\:{do}\:{i}\:{get}\:{to}\:{q}\mathrm{95473}? \\ $$
Answered by Sourav mridha last updated on 06/Jun/20
$$\boldsymbol{{ln}}\left(\mathrm{1}+\boldsymbol{{x}}\right)=\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\boldsymbol{{n}}}.\boldsymbol{{x}}^{\boldsymbol{{n}}} \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{putting}}\:\boldsymbol{{x}}=\mathrm{1}\:\boldsymbol{{you}}\:\boldsymbol{{get}}.. \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}−\mathrm{1}} }{\boldsymbol{{n}}}=\boldsymbol{{ln}}\mathrm{2}\approx\mathrm{0}.\mathrm{693}>\mathrm{0}\: \\ $$