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Question Number 93787 by Ar Brandon last updated on 14/May/20

Commented by mathmax by abdo last updated on 15/May/20

$$\frac{4k}{4k^4+1}=\frac{4k}{{\left(\sqrt{2}k\right)}^4+1}=\frac{4k}{{\left({\left(\sqrt{2}k\right)}^2\right)}^2+1}=\frac{4k}{{(2k^2+1)}^2-4k^2}$$$$=\frac{4k}{(2k^2+1-2k)(2k^2+1+2k)}=\frac{1}{(2k^2+1-2k)}-\frac{1}{2k^2+1+2k}$$$$\Rightarrow A=\sum _{k=1}^n(\frac{1}{2k^2-2k+1}-\frac{1}{2k^2+2k+1})$$$$let{u}_k=2k^2+2k+1\Rightarrow u_{k-1}=2{(k-1)}^2+2(k-1)+1$$$$=2(k^2-2k+1)+2k-2+1=2k^2-4k+2+2k-1$$$$=2k^2-2k+1\Rightarrow A=\sum _{k=1}^n(\frac{1}{u_{n-1}}-\frac{1}{u_n})$$$$A_n=\frac{1}{u_0}-\frac{1}{u_1}+\frac{1}{u_1}-\frac{1}{u_2}+....+\frac{1}{u_{n-1}}-\frac{1}{u_n}=\frac{1}{u_0}-\frac{1}{u_n}$$$$=1-\frac{1}{2n^2+2n+1}=\frac{2n^2+2n+1-1}{2n^2+2n+1}=\frac{2n^2+2n}{2n^2+2n+1}$$$$\Rightarrow A_n=\frac{2n^2+2n}{2n^2+2n+1}andweseethat{lim}_{n\to +\mathrm{\infty }}{A}_n=1$$

Commented by mathmax by abdo last updated on 15/May/20

$$A_n=\sum _{k=1}^n(\frac{1}{u_{k-1}}-\frac{1}{u_k})...$$

Commented by Ar Brandon last updated on 15/May/20

Great ��

Commented by mathmax by abdo last updated on 15/May/20

$$thanks$$

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