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1-1-2-1-3-1-4-1-5-1-6-1-7-Find-the-sum-




Question Number 99889 by Dwaipayan Shikari last updated on 23/Jun/20
1+(1/2)+(1/3)+(1/4)+(1/5)+(1/6)+(1/7)+.......∞{Find the sum}
$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{7}}+…….\infty\left\{\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\right\} \\ $$
Answered by smridha last updated on 23/Jun/20
this is called Harmonic series  and Harmonic series always  diverge so it has no finite value  H_n =Σ_(n=1) ^∞ (1/n)=∞  if the series is like that:  1−(1/2)+(1/3)−(1/4)+...∞=Σ_(n=1) ^∞ (((−1)^(n−1) )/n)(1)^n   =ln(2)≈0.693
$$\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{called}}\:\boldsymbol{{H}}{a}\boldsymbol{{rmonic}}\:\boldsymbol{{series}} \\ $$$$\boldsymbol{{and}}\:\boldsymbol{{H}}{a}\boldsymbol{{rmonic}}\:\boldsymbol{{series}}\:\boldsymbol{{always}} \\ $$$$\boldsymbol{{diverge}}\:\boldsymbol{{so}}\:\boldsymbol{{it}}\:\boldsymbol{{has}}\:\boldsymbol{{no}}\:\boldsymbol{{finite}}\:\boldsymbol{{value}} \\ $$$$\boldsymbol{{H}}_{\boldsymbol{{n}}} =\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{{n}}}=\infty \\ $$$$\boldsymbol{{if}}\:\boldsymbol{{the}}\:\boldsymbol{{series}}\:\boldsymbol{{is}}\:\boldsymbol{{like}}\:\boldsymbol{{that}}: \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+…\infty=\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\left(\mathrm{1}\right)^{\boldsymbol{{n}}} \\ $$$$=\boldsymbol{{ln}}\left(\mathrm{2}\right)\approx\mathrm{0}.\mathrm{693} \\ $$$$ \\ $$
Commented by Dwaipayan Shikari last updated on 23/Jun/20
Ohh  it is a logarithmic series .Thanking you
$${Ohh}\:\:{it}\:{is}\:{a}\:{logarithmic}\:{series}\:.{Thanking}\:{you} \\ $$

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