Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 172309 by mathocean1 last updated on 25/Jun/22

$$Determinate$$$$\underset{n\to +\mathrm{\infty }}{lim}\left(\sum _{k=1}^n\frac{{(-1)}^{k+1}}{k}\right)$$

Commented by mr W last updated on 25/Jun/22

$$\ln (1+x)=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{k+1}{x}^k}{k}$$$$withx=1:$$$$\ln 2=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{k+1}}{k}$$

Answered by Mathspace last updated on 25/Jun/22

$$u_n=\sum _{k=1}^n\frac{{(-1)}^{k-1}}{k}$$$$andp\left(x\right)=\sum _{k=1}^n\frac{{(-1)}^{k-1}}{k}{x}^k$$$$p^{{}^{\prime }}\left(x\right)=\sum _{k=1}^n{(-1)}^{k-1}{x}^{k-1}$$$$=\sum _{k=1}^n{(-x)}^{k-1}=\sum _{k=0}^{n-1}{(-x)}^k$$$$=\frac{1-{(-x)}^n}{1+x}\Rightarrow$$$$p\left(x\right)={\int }_0^x\frac{1-{(-t)}^n}{1+t}+c$$$$c=p\left(0\right)=0\Rightarrow$$$$p\left(x\right)={\int }_0^x\frac{1-{(-t)}^n}{1+t}dt$$$$\Rightarrow p\left(1\right)={\int }_0^1\frac{dt}{1+t}-{(-1)}^n{\int }_0^1\frac{t^n}{1+t}dt$$$${lim}_{n\to +\mathrm{\infty }}{u}_n={lim}_{n\to +\mathrm{\infty }}p\left(1\right)$$$$=ln2-{lim}_{n\to +\mathrm{\infty }}{(-1)}^n{\int }_0^1\frac{t^n}{1+t}dt$$$$=ln2-0=ln2$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com