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Question Number 42190 by maxmathsup by imad last updated on 19/Aug/18

$$let{u}_n=\sum _{k=1}^n\frac{{(-1)}^k}{\sqrt{k}}$$$$1)provethat\left(u_n\right)isconvergente$$$$2)findaequivalentof{u}_{n}whenn\to +\mathrm{\infty }$$

Commented by maxmathsup by imad last updated on 21/Aug/18

$$1)wehave{lim}_{n\to +\mathrm{\infty }}{u}_n=\sum _{k=1}^{\mathrm{\infty }}\frac{{(-1)}^k}{\sqrt{k}}andthisserieisconvergente$$$$\left(alternateserie\right)$$$$2)wehave{u}_n=\sum _{p=1}^{\left[\frac{n}{2}\right]}\frac{1}{\sqrt{2p}}-\sum _{p=0}^{\left[\frac{n-1}{2}\right]}\frac{1}{\sqrt{2p+1}}=\mathrm{\Sigma }{v}_p-\mathrm{\Sigma }{w}_p$$$$\left(v_p\right)isdecreasing\Rightarrow {\int }_p^{p+1}\frac{dx}{\sqrt{2x}}⩽v_p⩽{\int }_{p-1}^p\frac{dx}{\sqrt{2x}}\Rightarrow$$$$\sum _{p=1}^{\left[\frac{n}{2}\right]}{\int }_p^{p+1}\frac{dx}{\sqrt{2x}}⩽\sum _{p=1}^{\left[\frac{n}{2}\right]}{v}_p⩽\sum _{p=1}^{\left[\frac{n}{2}\right]}{\int }_{p-1}^p\frac{dx}{\sqrt{2x}}\Rightarrow$$$${\int }_1^{\left[\frac{n}{2}\right]+1}\frac{dx}{\sqrt{2x}}⩽\sum _{p=1}^{\left[\frac{n}{2}\right]}{v}_p⩽{\int }_0^{\left[\frac{n}{2}\right]}\frac{dx}{\sqrt{2x}}\Rightarrow$$$$\frac{1}{\sqrt{2}}{\left[2\sqrt{x}\right]}_1^{\left[\frac{n}{2}\right]+1}⩽\sum _{p=1}^{\left[\frac{n}{2}\right]}{v}_p⩽\frac{1}{\sqrt{2}}{\left[2\sqrt{x}\right]}_0^{\left[\frac{n}{2}\right]}\Rightarrow \sqrt{2}\{\sqrt{\left[\frac{n}{2}\right]+1}-1\}⩽\mathrm{\Sigma }{v}_p⩽\sqrt{2}\sqrt{\left[\frac{n}{2}\right]}$$$$also\left(w_p\right)isdecreasing\Rightarrow {\int }_p^{p+1}\frac{dx}{\sqrt{2x+1}}⩽w_p⩽{\int }_{p-1}^p\frac{dx}{\sqrt{2x+1}}\Rightarrow$$$$\sum _{p=1}^{\left[\frac{n-1}{2}\right]}{\int }_p^{p+1}\frac{dx}{\sqrt{2x+1}}⩽\sum _{p=1}^{\left[\frac{n-1}{2}\right]}{w}_p⩽\sum _{p=1}^{\left[\frac{n-1}{2}\right]}{\int }_{p-1}^p\frac{dx}{\sqrt{2x+1}}\Rightarrow$$$${\int }_1^{\left[\frac{n-1}{2}\right]+1}\frac{dx}{\sqrt{2x+1}}⩽\sum _{p=1}^{\left[\frac{n-1}{2}\right]}{w}_p⩽{\int }_0^{\left[\frac{n-1}{2}\right]}\frac{dx}{\sqrt{2x+1}}\Rightarrow$$$${\left[\sqrt{2x+1}\right]}_1^{\frac{[n-1}{2}]+1}⩽\sum _{p=1}^{\left[\frac{n-1}{2}\right]}{w}_p⩽{\left[\sqrt{2x+1}\right]}_0^{\left[\frac{n-1}{2}\right]}\Rightarrow$$$$\sqrt{2\left[\frac{n-1}{2}\right]+2}-\sqrt{3}⩽\sum _{p=1}^{\left[\frac{n-1}{2}\right]}{w}_p⩽\sqrt{2\left[\frac{n-1}{2}\right]+1}-1\Rightarrow$$$$\sqrt{2\left[\frac{n-1}{2}\right]+2}+1-\sqrt{3}⩽\sum _{p=0}^{\left[\frac{n-1}{2}\right]}{w}_p⩽\sqrt{2\left[\frac{n-1}{2}\right]+1}-2\Rightarrow$$$$\sqrt{2}\{\sqrt{\left[\frac{n}{2}\right]+1}-1\}+2-\sqrt{2\left[\frac{n-1}{2}\right]+1}⩽u_n⩽\sqrt{2}\sqrt{\left[\frac{n}{2}\right]}+\sqrt{3}-1-\sqrt{2\left[\frac{n-1}{2}\right]+1}$$$$...becontinued...$$$$$$

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