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# Summation question......Plz help

$$\sum_{i=1}^\infty (-1)^i$$ * $$\frac {1}{i}$$

3 years, 10 months ago

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$$\displaystyle \sum_{i=1}^{\infty} \dfrac{(-1)^i}{i} = -\log{2}$$ · 3 years, 10 months ago

can you please explain ? · 3 years, 10 months ago

$$\displaystyle \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+.....$$

Looks familiar? · 3 years, 10 months ago

Thanks. · 3 years, 10 months ago

by taylor series we have -log(1+y)= - y + (y^2)/2 + (y^3)/3........ which generates the above pattern by putting y=1 hence the answer for this is -log2 which would be around -0.30103 .. :) · 3 years, 10 months ago